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Operator overloading
is just “syntactic
sugar,” which means it is
simply another way for you to make a function call.
The difference is that the arguments for
this function don’t appear inside parentheses, but instead they surround
or are next to characters you’ve always thought of as immutable
operators.
There are two differences between the use
of an operator and an ordinary function call. The syntax is different; an
operator is often “called” by placing it between or sometimes after
the arguments. The second difference is that the compiler determines which
“function” to call. For instance, if you are using the operator
+ with floating-point arguments, the compiler “calls” the
function to perform floating-point addition (this “call” is
typically the act of inserting in-line code, or a floating-point-processor
instruction). If you use operator + with a floating-point number and an
integer, the compiler “calls” a special function to turn the
int into a float, and then “calls” the floating-point
addition code.
But in C++, it’s possible to define
new operators that work with classes. This definition is just like an ordinary
function definition except that the name of the function consists of the keyword
operator followed by the operator. That’s
the only difference, and it becomes a function like any
other function, which the compiler calls when it sees the appropriate
pattern.
It’s tempting to become
overenthusiastic with operator overloading. It’s a fun toy, at first. But
remember it’s only syntactic sugar, another way of calling a
function. Looking at it this way, you have no reason to overload an operator
except if it will make the code involving your class easier to write and
especially easier to read. (Remember, code is read much more than it is
written.) If this isn’t the case, don’t bother.
Another common response to operator
overloading is panic; suddenly, C operators have no familiar meaning anymore.
“Everything’s changed and all my C code will do different
things!” This isn’t true. All the operators used in expressions that
contain only built-in data types cannot be changed. You can never overload
operators such that
1 << 4;
behaves differently, or
1.414 << 2;
Defining an overloaded
operator is like defining a function, but the name of
that function is operator@, in which @ represents the operator
that’s being overloaded. The number of arguments in the overloaded
operator’s argument list depends on two
factors:
Here’s a
small class that shows the syntax for operator overloading:
//: C12:OperatorOverloadingSyntax.cpp
#include <iostream>
using namespace std;
class Integer {
int i;
public:
Integer(int ii) : i(ii) {}
const Integer
operator+(const Integer& rv) const {
cout << "operator+" << endl;
return Integer(i + rv.i);
}
Integer&
operator+=(const Integer& rv) {
cout << "operator+=" << endl;
i += rv.i;
return *this;
}
};
int main() {
cout << "built-in types:" << endl;
int i = 1, j = 2, k = 3;
k += i + j;
cout << "user-defined types:" << endl;
Integer ii(1), jj(2), kk(3);
kk += ii + jj;
} ///:~The two overloaded operators are defined
as inline member functions that announce when they are called. The single
argument is what appears on the right-hand side of the operator for binary
operators. Unary operators have no arguments when defined as member functions.
The member function is called for the object on the left-hand side of the
operator.
For non-conditional operators
(conditionals usually return a Boolean value), you’ll almost always want
to return an object or reference
of the same type you’re operating on if the two arguments are the same
type. (If they’re not the same type, the interpretation of what it should
produce is up to you.) This way,
complicated
expressions can be built up:
kk += ii + jj;
The operator+ produces a new
Integer (a temporary) that is used as the rv argument for the
operator+=. This temporary is destroyed as soon as it is no longer
needed.
Although you can overload
almost all the operators
available in C, the use of operator overloading is fairly restrictive. In
particular, you cannot combine operators that currently have no meaning in C
(such as ** to represent exponentiation), you cannot change the
evaluation precedence of operators, and you cannot change the number of
arguments required by an operator. This makes sense – all of these actions
would produce operators that confuse meaning rather than clarify
it.
The next two subsections give examples of
all the “regular” operators, overloaded in the form that
you’ll most likely
use.
The following example shows the syntax to
overload all the unary
operators, in the form of both
global functions (non-member friend functions) and as member functions.
These will expand upon the Integer class shown previously and add a new
byte class. The meaning of your particular operators will depend on the
way you want to use them, but consider the client programmer before doing
something unexpected.
Here is a catalog of all the unary
functions:
//: C12:OverloadingUnaryOperators.cpp
#include <iostream>
using namespace std;
// Non-member functions:
class Integer {
long i;
Integer* This() { return this; }
public:
Integer(long ll = 0) : i(ll) {}
// No side effects takes const& argument:
friend const Integer&
operator+(const Integer& a);
friend const Integer
operator-(const Integer& a);
friend const Integer
operator~(const Integer& a);
friend Integer*
operator&(Integer& a);
friend int
operator!(const Integer& a);
// Side effects have non-const& argument:
// Prefix:
friend const Integer&
operator++(Integer& a);
// Postfix:
friend const Integer
operator++(Integer& a, int);
// Prefix:
friend const Integer&
operator--(Integer& a);
// Postfix:
friend const Integer
operator--(Integer& a, int);
};
// Global operators:
const Integer& operator+(const Integer& a) {
cout << "+Integer\n";
return a; // Unary + has no effect
}
const Integer operator-(const Integer& a) {
cout << "-Integer\n";
return Integer(-a.i);
}
const Integer operator~(const Integer& a) {
cout << "~Integer\n";
return Integer(~a.i);
}
Integer* operator&(Integer& a) {
cout << "&Integer\n";
return a.This(); // &a is recursive!
}
int operator!(const Integer& a) {
cout << "!Integer\n";
return !a.i;
}
// Prefix; return incremented value
const Integer& operator++(Integer& a) {
cout << "++Integer\n";
a.i++;
return a;
}
// Postfix; return the value before increment:
const Integer operator++(Integer& a, int) {
cout << "Integer++\n";
Integer before(a.i);
a.i++;
return before;
}
// Prefix; return decremented value
const Integer& operator--(Integer& a) {
cout << "--Integer\n";
a.i--;
return a;
}
// Postfix; return the value before decrement:
const Integer operator--(Integer& a, int) {
cout << "Integer--\n";
Integer before(a.i);
a.i--;
return before;
}
// Show that the overloaded operators work:
void f(Integer a) {
+a;
-a;
~a;
Integer* ip = &a;
!a;
++a;
a++;
--a;
a--;
}
// Member functions (implicit "this"):
class Byte {
unsigned char b;
public:
Byte(unsigned char bb = 0) : b(bb) {}
// No side effects: const member function:
const Byte& operator+() const {
cout << "+Byte\n";
return *this;
}
const Byte operator-() const {
cout << "-Byte\n";
return Byte(-b);
}
const Byte operator~() const {
cout << "~Byte\n";
return Byte(~b);
}
Byte operator!() const {
cout << "!Byte\n";
return Byte(!b);
}
Byte* operator&() {
cout << "&Byte\n";
return this;
}
// Side effects: non-const member function:
const Byte& operator++() { // Prefix
cout << "++Byte\n";
b++;
return *this;
}
const Byte operator++(int) { // Postfix
cout << "Byte++\n";
Byte before(b);
b++;
return before;
}
const Byte& operator--() { // Prefix
cout << "--Byte\n";
--b;
return *this;
}
const Byte operator--(int) { // Postfix
cout << "Byte--\n";
Byte before(b);
--b;
return before;
}
};
void g(Byte b) {
+b;
-b;
~b;
Byte* bp = &b;
!b;
++b;
b++;
--b;
b--;
}
int main() {
Integer a;
f(a);
Byte b;
g(b);
} ///:~The functions are grouped according to
the way their arguments are passed. Guidelines for how to pass and return
arguments are given later. The forms above (and the ones that follow in the next
section) are typically what you’ll use, so start with them as a pattern
when overloading your own operators.
The overloaded ++ and –
– operators present a dilemma because you want to be able to call
different functions depending on whether they appear before (prefix) or after
(postfix) the object they’re acting upon. The solution is simple, but
people sometimes find it a bit confusing at first. When the compiler sees, for
example, ++a (a pre-increment), it generates a call to
operator++(a); but when
it sees a++, it generates a call to operator++(a, int). That is,
the compiler differentiates between the two forms by making calls to different
overloaded functions. In OverloadingUnaryOperators.cpp for the member
function versions, if the compiler sees ++b, it generates a call to
B::operator++( ); if it sees b++ it calls
B::operator++(int).
All the user sees is that a different
function gets called for the prefix
and
postfix versions. Underneath, however, the two functions calls have different
signatures, so they link to two different function bodies. The compiler passes a
dummy constant value for the int argument (which is never given an
identifier because the value is never used) to generate the different signature
for the postfix
version.
The following listing repeats the example
of OverloadingUnaryOperators.cpp for binary operators so you have an
example of all the operators you might want to overload. Again, both global
versions and member function versions are shown.
//: C12:Integer.h
// Non-member overloaded operators
#ifndef INTEGER_H
#define INTEGER_H
#include <iostream>
// Non-member functions:
class Integer {
long i;
public:
Integer(long ll = 0) : i(ll) {}
// Operators that create new, modified value:
friend const Integer
operator+(const Integer& left,
const Integer& right);
friend const Integer
operator-(const Integer& left,
const Integer& right);
friend const Integer
operator*(const Integer& left,
const Integer& right);
friend const Integer
operator/(const Integer& left,
const Integer& right);
friend const Integer
operator%(const Integer& left,
const Integer& right);
friend const Integer
operator^(const Integer& left,
const Integer& right);
friend const Integer
operator&(const Integer& left,
const Integer& right);
friend const Integer
operator|(const Integer& left,
const Integer& right);
friend const Integer
operator<<(const Integer& left,
const Integer& right);
friend const Integer
operator>>(const Integer& left,
const Integer& right);
// Assignments modify & return lvalue:
friend Integer&
operator+=(Integer& left,
const Integer& right);
friend Integer&
operator-=(Integer& left,
const Integer& right);
friend Integer&
operator*=(Integer& left,
const Integer& right);
friend Integer&
operator/=(Integer& left,
const Integer& right);
friend Integer&
operator%=(Integer& left,
const Integer& right);
friend Integer&
operator^=(Integer& left,
const Integer& right);
friend Integer&
operator&=(Integer& left,
const Integer& right);
friend Integer&
operator|=(Integer& left,
const Integer& right);
friend Integer&
operator>>=(Integer& left,
const Integer& right);
friend Integer&
operator<<=(Integer& left,
const Integer& right);
// Conditional operators return true/false:
friend int
operator==(const Integer& left,
const Integer& right);
friend int
operator!=(const Integer& left,
const Integer& right);
friend int
operator<(const Integer& left,
const Integer& right);
friend int
operator>(const Integer& left,
const Integer& right);
friend int
operator<=(const Integer& left,
const Integer& right);
friend int
operator>=(const Integer& left,
const Integer& right);
friend int
operator&&(const Integer& left,
const Integer& right);
friend int
operator||(const Integer& left,
const Integer& right);
// Write the contents to an ostream:
void print(std::ostream& os) const { os << i; }
};
#endif // INTEGER_H ///:~//: C12:Integer.cpp {O}
// Implementation of overloaded operators
#include "Integer.h"
#include "../require.h"
const Integer
operator+(const Integer& left,
const Integer& right) {
return Integer(left.i + right.i);
}
const Integer
operator-(const Integer& left,
const Integer& right) {
return Integer(left.i - right.i);
}
const Integer
operator*(const Integer& left,
const Integer& right) {
return Integer(left.i * right.i);
}
const Integer
operator/(const Integer& left,
const Integer& right) {
require(right.i != 0, "divide by zero");
return Integer(left.i / right.i);
}
const Integer
operator%(const Integer& left,
const Integer& right) {
require(right.i != 0, "modulo by zero");
return Integer(left.i % right.i);
}
const Integer
operator^(const Integer& left,
const Integer& right) {
return Integer(left.i ^ right.i);
}
const Integer
operator&(const Integer& left,
const Integer& right) {
return Integer(left.i & right.i);
}
const Integer
operator|(const Integer& left,
const Integer& right) {
return Integer(left.i | right.i);
}
const Integer
operator<<(const Integer& left,
const Integer& right) {
return Integer(left.i << right.i);
}
const Integer
operator>>(const Integer& left,
const Integer& right) {
return Integer(left.i >> right.i);
}
// Assignments modify & return lvalue:
Integer& operator+=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i += right.i;
return left;
}
Integer& operator-=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i -= right.i;
return left;
}
Integer& operator*=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i *= right.i;
return left;
}
Integer& operator/=(Integer& left,
const Integer& right) {
require(right.i != 0, "divide by zero");
if(&left == &right) {/* self-assignment */}
left.i /= right.i;
return left;
}
Integer& operator%=(Integer& left,
const Integer& right) {
require(right.i != 0, "modulo by zero");
if(&left == &right) {/* self-assignment */}
left.i %= right.i;
return left;
}
Integer& operator^=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i ^= right.i;
return left;
}
Integer& operator&=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i &= right.i;
return left;
}
Integer& operator|=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i |= right.i;
return left;
}
Integer& operator>>=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i >>= right.i;
return left;
}
Integer& operator<<=(Integer& left,
const Integer& right) {
if(&left == &right) {/* self-assignment */}
left.i <<= right.i;
return left;
}
// Conditional operators return true/false:
int operator==(const Integer& left,
const Integer& right) {
return left.i == right.i;
}
int operator!=(const Integer& left,
const Integer& right) {
return left.i != right.i;
}
int operator<(const Integer& left,
const Integer& right) {
return left.i < right.i;
}
int operator>(const Integer& left,
const Integer& right) {
return left.i > right.i;
}
int operator<=(const Integer& left,
const Integer& right) {
return left.i <= right.i;
}
int operator>=(const Integer& left,
const Integer& right) {
return left.i >= right.i;
}
int operator&&(const Integer& left,
const Integer& right) {
return left.i && right.i;
}
int operator||(const Integer& left,
const Integer& right) {
return left.i || right.i;
} ///:~//: C12:IntegerTest.cpp
//{L} Integer
#include "Integer.h"
#include <fstream>
using namespace std;
ofstream out("IntegerTest.out");
void h(Integer& c1, Integer& c2) {
// A complex expression:
c1 += c1 * c2 + c2 % c1;
#define TRY(OP) \
out << "c1 = "; c1.print(out); \
out << ", c2 = "; c2.print(out); \
out << "; c1 " #OP " c2 produces "; \
(c1 OP c2).print(out); \
out << endl;
TRY(+) TRY(-) TRY(*) TRY(/)
TRY(%) TRY(^) TRY(&) TRY(|)
TRY(<<) TRY(>>) TRY(+=) TRY(-=)
TRY(*=) TRY(/=) TRY(%=) TRY(^=)
TRY(&=) TRY(|=) TRY(>>=) TRY(<<=)
// Conditionals:
#define TRYC(OP) \
out << "c1 = "; c1.print(out); \
out << ", c2 = "; c2.print(out); \
out << "; c1 " #OP " c2 produces "; \
out << (c1 OP c2); \
out << endl;
TRYC(<) TRYC(>) TRYC(==) TRYC(!=) TRYC(<=)
TRYC(>=) TRYC(&&) TRYC(||)
}
int main() {
cout << "friend functions" << endl;
Integer c1(47), c2(9);
h(c1, c2);
} ///:~//: C12:Byte.h
// Member overloaded operators
#ifndef BYTE_H
#define BYTE_H
#include "../require.h"
#include <iostream>
// Member functions (implicit "this"):
class Byte {
unsigned char b;
public:
Byte(unsigned char bb = 0) : b(bb) {}
// No side effects: const member function:
const Byte
operator+(const Byte& right) const {
return Byte(b + right.b);
}
const Byte
operator-(const Byte& right) const {
return Byte(b - right.b);
}
const Byte
operator*(const Byte& right) const {
return Byte(b * right.b);
}
const Byte
operator/(const Byte& right) const {
require(right.b != 0, "divide by zero");
return Byte(b / right.b);
}
const Byte
operator%(const Byte& right) const {
require(right.b != 0, "modulo by zero");
return Byte(b % right.b);
}
const Byte
operator^(const Byte& right) const {
return Byte(b ^ right.b);
}
const Byte
operator&(const Byte& right) const {
return Byte(b & right.b);
}
const Byte
operator|(const Byte& right) const {
return Byte(b | right.b);
}
const Byte
operator<<(const Byte& right) const {
return Byte(b << right.b);
}
const Byte
operator>>(const Byte& right) const {
return Byte(b >> right.b);
}
// Assignments modify & return lvalue.
// operator= can only be a member function:
Byte& operator=(const Byte& right) {
// Handle self-assignment:
if(this == &right) return *this;
b = right.b;
return *this;
}
Byte& operator+=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b += right.b;
return *this;
}
Byte& operator-=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b -= right.b;
return *this;
}
Byte& operator*=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b *= right.b;
return *this;
}
Byte& operator/=(const Byte& right) {
require(right.b != 0, "divide by zero");
if(this == &right) {/* self-assignment */}
b /= right.b;
return *this;
}
Byte& operator%=(const Byte& right) {
require(right.b != 0, "modulo by zero");
if(this == &right) {/* self-assignment */}
b %= right.b;
return *this;
}
Byte& operator^=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b ^= right.b;
return *this;
}
Byte& operator&=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b &= right.b;
return *this;
}
Byte& operator|=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b |= right.b;
return *this;
}
Byte& operator>>=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b >>= right.b;
return *this;
}
Byte& operator<<=(const Byte& right) {
if(this == &right) {/* self-assignment */}
b <<= right.b;
return *this;
}
// Conditional operators return true/false:
int operator==(const Byte& right) const {
return b == right.b;
}
int operator!=(const Byte& right) const {
return b != right.b;
}
int operator<(const Byte& right) const {
return b < right.b;
}
int operator>(const Byte& right) const {
return b > right.b;
}
int operator<=(const Byte& right) const {
return b <= right.b;
}
int operator>=(const Byte& right) const {
return b >= right.b;
}
int operator&&(const Byte& right) const {
return b && right.b;
}
int operator||(const Byte& right) const {
return b || right.b;
}
// Write the contents to an ostream:
void print(std::ostream& os) const {
os << "0x" << std::hex << int(b) << std::dec;
}
};
#endif // BYTE_H ///:~//: C12:ByteTest.cpp
#include "Byte.h"
#include <fstream>
using namespace std;
ofstream out("ByteTest.out");
void k(Byte& b1, Byte& b2) {
b1 = b1 * b2 + b2 % b1;
#define TRY2(OP) \
out << "b1 = "; b1.print(out); \
out << ", b2 = "; b2.print(out); \
out << "; b1 " #OP " b2 produces "; \
(b1 OP b2).print(out); \
out << endl;
b1 = 9; b2 = 47;
TRY2(+) TRY2(-) TRY2(*) TRY2(/)
TRY2(%) TRY2(^) TRY2(&) TRY2(|)
TRY2(<<) TRY2(>>) TRY2(+=) TRY2(-=)
TRY2(*=) TRY2(/=) TRY2(%=) TRY2(^=)
TRY2(&=) TRY2(|=) TRY2(>>=) TRY2(<<=)
TRY2(=) // Assignment operator
// Conditionals:
#define TRYC2(OP) \
out << "b1 = "; b1.print(out); \
out << ", b2 = "; b2.print(out); \
out << "; b1 " #OP " b2 produces "; \
out << (b1 OP b2); \
out << endl;
b1 = 9; b2 = 47;
TRYC2(<) TRYC2(>) TRYC2(==) TRYC2(!=) TRYC2(<=)
TRYC2(>=) TRYC2(&&) TRYC2(||)
// Chained assignment:
Byte b3 = 92;
b1 = b2 = b3;
}
int main() {
out << "member functions:" << endl;
Byte b1(47), b2(9);
k(b1, b2);
} ///:~You can see that operator= is only
allowed to be a member function. This is explained later.
Notice that all of the assignment
operators have code to check for
self-assignment;
this is a general guideline. In some cases this is not necessary; for example,
with operator+= you often want to say A+=A and have it add
A to itself. The most important place to check for self-assignment is
operator= because with
complicated objects disastrous results may occur. (In some cases it’s OK,
but you should always keep it in mind when writing
operator=.)
All of the operators shown in the
previous two examples are overloaded to handle a single type. It’s also
possible to overload operators to handle mixed types, so you can add apples to
oranges, for example. Before you start on an exhaustive overloading of
operators, however, you should look at the section on automatic type conversion
later in this chapter. Often, a type conversion in the right place can save you
a lot of overloaded
operators.
It may seem a little confusing at first
when you look at OverloadingUnaryOperators.cpp, Integer.h and
Byte.h and see all the different ways that arguments are passed and
returned. Although you can pass and return arguments any way you want to,
the choices in these examples were not selected at random. They follow a logical
pattern, the same one you’ll want to use in most of your
choices.
The increment
and decrement operators
present
a dilemma because of the pre- and postfix versions. Both versions change the
object and so cannot treat the object as a const. The prefix version
returns the value of the object after it was changed, so you expect to get back
the object that was changed. Thus, with prefix you can just return *this
as a reference. The postfix version is supposed to return the value
before the value is changed, so you’re forced to create a separate
object to represent that value and return it. So with postfix you must return by
value if you want to preserve the expected meaning. (Note that you’ll
sometimes find the increment and decrement operators returning an int or
bool to indicate, for example, whether an object designed to move through
a list is at the end of that list.) Now the question is: Should these be
returned as const or nonconst? If you allow the object to be
modified and someone writes (++a).func( ), func( ) will
be operating on a itself, but with (a++).func( ),
func( ) operates on the temporary object returned by the postfix
operator++. Temporary objects are automatically const, so this
would be flagged by the compiler, but for consistency’s sake it may make
more sense to make them both const, as was done here. Or you may choose
to make the prefix version non-const and the postfix const.
Because of the variety of meanings you may want to give the increment and
decrement operators, they will need to be considered on a case-by-case
basis.
Returning by value as a const can
seem a bit subtle at first, so it deserves a bit more explanation. Consider the
binary operator+. If you use it in an expression such as f(a+b),
the result of a+b becomes a temporary object that is used in the call to
f( ). Because it’s a temporary, it’s automatically
const, so whether you explicitly make the return value const or
not has no effect.
However, it’s also possible for you
to send a message to the return value of a+b, rather than just passing it
to a function. For example, you can say (a+b).g( ), in which
g( ) is some member function of Integer, in this case. By
making the return value const, you state that only a const member
function can be called for that return value. This is const-correct,
because it prevents you from storing potentially valuable information in an
object that will most likely be lost.
When new objects are created to return by
value, notice the form used. In operator+, for example:
return Integer(left.i + right.i);
This may look at first like a
“function call to a constructor,” but it’s not. The syntax is
that of a temporary object; the statement says “make a temporary
Integer object and return it.” Because of this, you might think
that the result is the same as creating a named local object and returning that.
However, it’s quite different. If you were to say
instead:
Integer tmp(left.i + right.i); return tmp;
three things will happen. First, the
tmp object is created including its constructor call. Second, the
copy-constructor copies the tmp to the location
of the outside return value. Third, the destructor is called for tmp at
the end of the scope.
In contrast, the “returning a
temporary” approach works quite differently. When
the compiler sees you do this, it knows that you have no other need for the
object it’s creating than to return it. The compiler takes advantage of
this by building the object directly into the location of the outside
return value. This requires only a single ordinary constructor call (no
copy-constructor is necessary) and there’s no destructor call because you
never actually create a local object. Thus, while it doesn’t cost anything
but programmer awareness, it’s significantly more efficient. This is often
called the return value
optimization.
Several additional operators have a
slightly different syntax for overloading.
The subscript, operator[
], must be a member function and it requires a
single argument. Because operator[ ] implies that the object it’s
being called for acts like an array, you will often return a reference from this
operator, so it can be conveniently used on the left-hand side of an equal sign.
This operator is commonly overloaded; you’ll see examples in the rest of
the book.
The operators new and
delete control dynamic storage allocation and can be overloaded in a
number of different ways. This topic is covered in the Chapter
13.
The comma
operator is called when it
appears next to an object of the type the comma is defined for. However,
“operator,” is not called for function argument
lists, only for objects that are out in the open, separated by commas. There
doesn’t seem to be a lot of practical uses for this operator; it’s
in the language for consistency. Here’s an example showing how the comma
function can be called when the comma appears before an object, as well
as after:
//: C12:OverloadingOperatorComma.cpp
#include <iostream>
using namespace std;
class After {
public:
const After& operator,(const After&) const {
cout << "After::operator,()" << endl;
return *this;
}
};
class Before {};
Before& operator,(int, Before& b) {
cout << "Before::operator,()" << endl;
return b;
}
int main() {
After a, b;
a, b; // Operator comma called
Before c;
1, c; // Operator comma called
} ///:~The global function allows the comma to
be placed before the object in question. The usage shown is fairly obscure and
questionable. Although you would probably use a comma-separated list as part of
a more complex expression, it’s too subtle to use in most
situations.
The
operator–>
is generally used when you want to make an object appear to be a pointer. Since
such an object has more “smarts” built into it than exist for a
typical pointer, an object like this is often called a smart pointer.
These are especially useful if you want to “wrap” a class around a
pointer to make that pointer safe, or in the common usage of an
iterator, which is an object that moves through a
collection /container
of other objects and selects them one at a time,
without providing direct access to the implementation of the container.
(You’ll often find containers and iterators in class libraries, such as in
the Standard C++ Library, described in Volume 2 of this book.)
A pointer dereference operator must be a
member function. It has additional, atypical constraints: It must return an
object (or reference to an object) that also has a pointer dereference operator,
or it must return a pointer that can be used to select what the pointer
dereference operator arrow is pointing at. Here’s a simple
example:
//: C12:SmartPointer.cpp
#include <iostream>
#include <vector>
#include "../require.h"
using namespace std;
class Obj {
static int i, j;
public:
void f() const { cout << i++ << endl; }
void g() const { cout << j++ << endl; }
};
// Static member definitions:
int Obj::i = 47;
int Obj::j = 11;
// Container:
class ObjContainer {
vector<Obj*> a;
public:
void add(Obj* obj) { a.push_back(obj); }
friend class SmartPointer;
};
class SmartPointer {
ObjContainer& oc;
int index;
public:
SmartPointer(ObjContainer& objc) : oc(objc) {
index = 0;
}
// Return value indicates end of list:
bool operator++() { // Prefix
if(index >= oc.a.size()) return false;
if(oc.a[++index] == 0) return false;
return true;
}
bool operator++(int) { // Postfix
return operator++(); // Use prefix version
}
Obj* operator->() const {
require(oc.a[index] != 0, "Zero value "
"returned by SmartPointer::operator->()");
return oc.a[index];
}
};
int main() {
const int sz = 10;
Obj o[sz];
ObjContainer oc;
for(int i = 0; i < sz; i++)
oc.add(&o[i]); // Fill it up
SmartPointer sp(oc); // Create an iterator
do {
sp->f(); // Pointer dereference operator call
sp->g();
} while(sp++);
} ///:~The class Obj defines the objects
that are manipulated in this program. The functions f( ) and
g( ) simply print out interesting values using static data
members. Pointers to these objects are stored inside containers of type
ObjContainer using its add( ) function. ObjContainer
looks like an array of pointers, but you’ll notice there’s no way to
get the pointers back out again. However, SmartPointer is declared as a
friend class, so it has permission to look inside the container. The
SmartPointer class looks very much like an intelligent pointer –
you can move it forward using operator++ (you can also define an
operator– –), it won’t go past the end of the container
it’s pointing to, and it produces (via the pointer dereference operator)
the value it’s pointing to. Notice that the SmartPointer is a
custom fit for the container it’s created for; unlike an ordinary pointer,
there isn’t a “general purpose” smart pointer. You will learn
more about the smart pointers called “iterators” in the last chapter
of this book and in Volume 2 (downloadable from
www.BruceEckel.com).
In main( ), once the
container oc is filled with Obj objects, a SmartPointer sp
is created. The smart pointer calls happen in the expressions:
sp->f(); // Smart pointer calls sp->g();
Here, even though sp doesn’t
actually have f( ) and g( ) member functions, the
pointer dereference operator automatically calls those functions for the
Obj* that is returned by SmartPointer::operator–>. The
compiler performs all the checking to make sure the function call works
properly.
Although the underlying mechanics of the
pointer dereference operator are more complex than the other operators, the goal
is exactly the same: to provide a more convenient syntax for the users of your
classes.
It’s more common to see a
“smart pointer” or “iterator” class nested within the
class that it services. The previous example can be rewritten to nest
SmartPointer inside ObjContainer like this:
//: C12:NestedSmartPointer.cpp
#include <iostream>
#include <vector>
#include "../require.h"
using namespace std;
class Obj {
static int i, j;
public:
void f() { cout << i++ << endl; }
void g() { cout << j++ << endl; }
};
// Static member definitions:
int Obj::i = 47;
int Obj::j = 11;
// Container:
class ObjContainer {
vector<Obj*> a;
public:
void add(Obj* obj) { a.push_back(obj); }
class SmartPointer;
friend class SmartPointer;
class SmartPointer {
ObjContainer& oc;
unsigned int index;
public:
SmartPointer(ObjContainer& objc) : oc(objc) {
index = 0;
}
// Return value indicates end of list:
bool operator++() { // Prefix
if(index >= oc.a.size()) return false;
if(oc.a[++index] == 0) return false;
return true;
}
bool operator++(int) { // Postfix
return operator++(); // Use prefix version
}
Obj* operator->() const {
require(oc.a[index] != 0, "Zero value "
"returned by SmartPointer::operator->()");
return oc.a[index];
}
};
// Function to produce a smart pointer that
// points to the beginning of the ObjContainer:
SmartPointer begin() {
return SmartPointer(*this);
}
};
int main() {
const int sz = 10;
Obj o[sz];
ObjContainer oc;
for(int i = 0; i < sz; i++)
oc.add(&o[i]); // Fill it up
ObjContainer::SmartPointer sp = oc.begin();
do {
sp->f(); // Pointer dereference operator call
sp->g();
} while(++sp);
} ///:~Besides the actual nesting of the class,
there are only two differences here. The first is in the
declaration
of the class so that it can be a friend:
class SmartPointer; friend SmartPointer;
The compiler must first know that the
class exists before it can be told that it’s a
friend.
The second difference is in the
ObjContainer member function begin( ), which produces a
SmartPointer that points to the beginning of the ObjContainer
sequence. Although it’s really only a convenience, it’s valuable
because it follows part of the form used in the Standard C++
Library.
The operator–>*
is a binary operator that
behaves like all the other binary operators. It is provided for those situations
when you want to mimic the behavior provided by the built-in
pointer-to-member syntax, described in the
previous chapter.
Just like operator->, the
pointer-to-member dereference operator is generally used with some kind of
object that represents a “smart pointer,” although the example shown
here will be simpler so it’s understandable. The trick when defining
operator->* is that it must return an object for which the
operator( ) can be called with the arguments for the member function
you’re calling.
The function call
operator( )
must be a member function, and it is unique in that it allows any number of
arguments. It makes your object look like it’s actually a function.
Although you could define several overloaded operator( ) functions
with different arguments, it’s often used for types that only have a
single operation, or at least an especially prominent one. You’ll see in
Volume 2 that the Standard C++ Library uses the function call operator in order
to create “function objects.”
To create an operator->* you
must first create a class with an operator( ) that is the type of
object that operator->* will return. This class must somehow capture
the necessary information so that when the operator( ) is called
(which happens automatically), the pointer-to-member will be dereferenced for
the object. In the following example, the FunctionObject constructor
captures and stores both the pointer to the object and the pointer to the member
function, and then the operator( ) uses those to make the actual
pointer-to-member call:
//: C12:PointerToMemberOperator.cpp
#include <iostream>
using namespace std;
class Dog {
public:
int run(int i) const {
cout << "run\n";
return i;
}
int eat(int i) const {
cout << "eat\n";
return i;
}
int sleep(int i) const {
cout << "ZZZ\n";
return i;
}
typedef int (Dog::*PMF)(int) const;
// operator->* must return an object
// that has an operator():
class FunctionObject {
Dog* ptr;
PMF pmem;
public:
// Save the object pointer and member pointer
FunctionObject(Dog* wp, PMF pmf)
: ptr(wp), pmem(pmf) {
cout << "FunctionObject constructor\n";
}
// Make the call using the object pointer
// and member pointer
int operator()(int i) const {
cout << "FunctionObject::operator()\n";
return (ptr->*pmem)(i); // Make the call
}
};
FunctionObject operator->*(PMF pmf) {
cout << "operator->*" << endl;
return FunctionObject(this, pmf);
}
};
int main() {
Dog w;
Dog::PMF pmf = &Dog::run;
cout << (w->*pmf)(1) << endl;
pmf = &Dog::sleep;
cout << (w->*pmf)(2) << endl;
pmf = &Dog::eat;
cout << (w->*pmf)(3) << endl;
} ///:~Dog has three member functions,
all of which take an int argument and return an int. PMF is
a typedef to simplify defining a pointer-to-member to Dog’s
member functions.
A FunctionObject is created and
returned by operator->*. Notice that operator->* knows both
the object that the pointer-to-member is being called for (this) and the
pointer-to-member, and it passes those to the FunctionObject constructor
that stores the values. When operator->* is called, the compiler
immediately turns around and calls operator( ) for the return value
of operator->*, passing in the arguments that were given to
operator->*. The FunctionObject::operator( ) takes the
arguments and then dereferences the “real” pointer-to-member using
its stored object pointer and pointer-to-member.
Notice that what you are doing here, just
as with operator->, is inserting yourself in the middle of the call to
operator->*. This allows you to perform some extra operations if you
need to.
The operator->* mechanism
implemented here only works for member functions that take an int
argument and return an int. This is limiting, but if you try to
create overloaded mechanisms for each different possibility, it seems like a
prohibitive task. Fortunately, C++’s template mechanism (described
in the last chapter of this book, and in Volume 2) is designed to handle just
such a
problem.
There are certain operators in the
available set that cannot be overloaded. The general reason for the restriction
is safety. If these operators were overloadable, it would somehow jeopardize or
break safety mechanisms, make things harder, or confuse existing
practice.
In some of the previous examples, the
operators may be members or non-members, and it doesn’t seem to make much
difference. This usually raises the question, “Which should I
choose?” In general, if it doesn’t make any difference, they should
be members, to emphasize the association between the operator and its class.
When the left-hand operand is always an object of the current class, this works
fine.
However, sometimes you want the left-hand
operand to be an object of some other class. A common place you’ll see
this is when the operators << and >> are overloaded
for
iostreams.
Since iostreams is a fundamental C++ library, you’ll probably want to
overload these operators for most of your classes, so the process is worth
memorizing:
//: C12:IostreamOperatorOverloading.cpp
// Example of non-member overloaded operators
#include "../require.h"
#include <iostream>
#include <sstream> // "String streams"
#include <cstring>
using namespace std;
class IntArray {
enum { sz = 5 };
int i[sz];
public:
IntArray() { memset(i, 0, sz* sizeof(*i)); }
int& operator[](int x) {
require(x >= 0 && x < sz,
"IntArray::operator[] out of range");
return i[x];
}
friend ostream&
operator<<(ostream& os, const IntArray& ia);
friend istream&
operator>>(istream& is, IntArray& ia);
};
ostream&
operator<<(ostream& os, const IntArray& ia) {
for(int j = 0; j < ia.sz; j++) {
os << ia.i[j];
if(j != ia.sz -1)
os << ", ";
}
os << endl;
return os;
}
istream& operator>>(istream& is, IntArray& ia){
for(int j = 0; j < ia.sz; j++)
is >> ia.i[j];
return is;
}
int main() {
stringstream input("47 34 56 92 103");
IntArray I;
input >> I;
I[4] = -1; // Use overloaded operator[]
cout << I;
} ///:~This class also contains an overloaded
operator
[ ], which returns a reference to a legitimate value in the array. Because a
reference is returned, the expression
I[4] = -1;
not only looks much more civilized than
if pointers were used, it also accomplishes the desired effect.
It’s important that the overloaded
shift operators pass and return by reference, so the actions will affect
the external objects. In the function definitions, expressions
like
os << ia.i[j];
cause the existing overloaded
operator functions to be called (that is, those defined in
<iostream>). In this case, the function called is ostream&
operator<<(ostream&, int) because ia.i[j] resolves to an
int.
Once all the actions are performed on the
istream or
ostream, it is returned so it can be used in a
more complicated expression.
In main( ), a new type of
iostream is used: the stringstream
(declared in
<sstream>). This is
a class that takes a string (which it can create from a char
array, as shown here) and turns it into an iostream. In the example
above, this means that the shift operators can be tested without opening a file
or typing data in on the command line.
The form shown in this example for the
inserter and extractor is standard. If you want to create these operators for
your own class, copy the function signatures and return types above and follow
the form of the
body.
Murray[49]
suggests these guidelines for choosing between members and
non-members:
|
Operator
|
Recommended use
|
|---|---|
|
All unary operators |
member |
|
= ( ) [ ] –>
–>* |
must be member |
|
+= –= /= *= ^=
|
member |
|
All other binary
operators |
non-member |
A common source of confusion with new C++
programmers is assignment. This is no doubt because the = sign is such a
fundamental operation in programming, right down to copying a register at the
machine level. In addition, the
copy-constructor
(described in Chapter 11) is also sometimes invoked when the
= sign is
used:
MyType b; MyType a = b; a = b;
In the second line, the object a
is being defined. A new object is being created where one didn’t
exist before. Because you know by now how defensive the C++ compiler is about
object initialization, you know that a constructor must always be called at the
point where an object is defined. But which constructor? a is being
created from an existing MyType object (b, on the right side of
the equal sign), so there’s only one choice: the copy-constructor. Even
though an equal sign is involved, the copy-constructor is
called.
In the third line, things are different.
On the left side of the equal sign, there’s a previously initialized
object. Clearly, you don’t call a constructor for an object that’s
already been created. In this case MyType::operator= is called for
a, taking as an argument whatever appears on the right-hand side. (You
can have multiple operator= functions to take different types of
right-hand arguments.)
This behavior is not restricted to the
copy-constructor. Any time you’re initializing an
object using an = instead of the ordinary function-call form of the
constructor, the compiler will look for a constructor that accepts whatever is
on the right-hand side:
//: C12:CopyingVsInitialization.cpp
class Fi {
public:
Fi() {}
};
class Fee {
public:
Fee(int) {}
Fee(const Fi&) {}
};
int main() {
Fee fee = 1; // Fee(int)
Fi fi;
Fee fum = fi; // Fee(Fi)
} ///:~When dealing with the = sign,
it’s important to keep this distinction in mind: If the object
hasn’t been created yet, initialization is required; otherwise the
assignment operator= is used.
It’s even better to avoid writing
code that uses the = for initialization; instead, always use the explicit
constructor form. The two constructions with the equal sign then
become:
Fee fee(1); Fee fum(fi);
In Integer.h and Byte.h,
you saw that operator= can be only a member function. It is intimately
connected to the object on the left side of the ‘=’. If it
was possible to define operator= globally, then you might attempt to
redefine the built-in ‘=’ sign:
int operator=(int, MyType); // Global = not allowed!
The compiler skirts this whole issue by
forcing you to make operator= a member function.
When you create an operator=, you
must copy all of the necessary information from the right-hand object into the
current object (that is, the object that operator= is being called for)
to perform whatever you consider “assignment” for your class. For
simple objects, this is obvious:
//: C12:SimpleAssignment.cpp
// Simple operator=()
#include <iostream>
using namespace std;
class Value {
int a, b;
float c;
public:
Value(int aa = 0, int bb = 0, float cc = 0.0)
: a(aa), b(bb), c(cc) {}
Value& operator=(const Value& rv) {
a = rv.a;
b = rv.b;
c = rv.c;
return *this;
}
friend ostream&
operator<<(ostream& os, const Value& rv) {
return os << "a = " << rv.a << ", b = "
<< rv.b << ", c = " << rv.c;
}
};
int main() {
Value a, b(1, 2, 3.3);
cout << "a: " << a << endl;
cout << "b: " << b << endl;
a = b;
cout << "a after assignment: " << a << endl;
} ///:~Here, the object on the left side of the
= copies all the elements of the object on the right, then returns a
reference to itself, which allows a more complex expression to be
created.
This example includes a common mistake.
When you’re assigning two objects of the same type, you should always
check first for self-assignment:
is the object being assigned to itself? In some cases, such as this one,
it’s harmless if you perform the assignment operations anyway, but if
changes are made to the implementation of the class, it can make a difference,
and if you don’t do it as a matter of habit, you may forget and cause
hard-to-find bugs.
What happens if the object is not so
simple? For example, what if the object contains pointers to other objects?
Simply copying a pointer means that you’ll end up
with two objects pointing to the same storage location. In situations like
these, you need to do bookkeeping of your own.
There are two common approaches to this
problem. The simplest technique is to copy whatever the pointer refers to when
you do an assignment or a copy-construction. This is
straightforward:
//: C12:CopyingWithPointers.cpp
// Solving the pointer aliasing problem by
// duplicating what is pointed to during
// assignment and copy-construction.
#include "../require.h"
#include <string>
#include <iostream>
using namespace std;
class Dog {
string nm;
public:
Dog(const string& name) : nm(name) {
cout << "Creating Dog: " << *this << endl;
}
// Synthesized copy-constructor & operator=
// are correct.
// Create a Dog from a Dog pointer:
Dog(const Dog* dp, const string& msg)
: nm(dp->nm + msg) {
cout << "Copied dog " << *this << " from "
<< *dp << endl;
}
~Dog() {
cout << "Deleting Dog: " << *this << endl;
}
void rename(const string& newName) {
nm = newName;
cout << "Dog renamed to: " << *this << endl;
}
friend ostream&
operator<<(ostream& os, const Dog& d) {
return os << "[" << d.nm << "]";
}
};
class DogHouse {
Dog* p;
string houseName;
public:
DogHouse(Dog* dog, const string& house)
: p(dog), houseName(house) {}
DogHouse(const DogHouse& dh)
: p(new Dog(dh.p, " copy-constructed")),
houseName(dh.houseName
+ " copy-constructed") {}
DogHouse& operator=(const DogHouse& dh) {
// Check for self-assignment:
if(&dh != this) {
p = new Dog(dh.p, " assigned");
houseName = dh.houseName + " assigned";
}
return *this;
}
void renameHouse(const string& newName) {
houseName = newName;
}
Dog* getDog() const { return p; }
~DogHouse() { delete p; }
friend ostream&
operator<<(ostream& os, const DogHouse& dh) {
return os << "[" << dh.houseName
<< "] contains " << *dh.p;
}
};
int main() {
DogHouse fidos(new Dog("Fido"), "FidoHouse");
cout << fidos << endl;
DogHouse fidos2 = fidos; // Copy construction
cout << fidos2 << endl;
fidos2.getDog()->rename("Spot");
fidos2.renameHouse("SpotHouse");
cout << fidos2 << endl;
fidos = fidos2; // Assignment
cout << fidos << endl;
fidos.getDog()->rename("Max");
fidos2.renameHouse("MaxHouse");
} ///:~Dog is a simple class that
contains only a string that holds the name of the dog. However,
you’ll generally know when something happens to a Dog because the
constructors and destructors print information when they are called. Notice that
the second constructor is a bit like a copy-constructor except that it takes a
pointer to a Dog instead of a reference, and it has a second argument
that is a message that’s concatenated to the argument Dog’s
name. This is used to help trace the behavior of the program.
You can see that whenever a member
function prints information, it doesn’t access that information directly
but instead sends *this to cout. This in turn calls the
ostream operator<<. It’s valuable to do it this way
because if you want to reformat the way that Dog information is displayed
(as I did by adding the ‘[’ and ‘]’) you only need to do
it in one place.
A DogHouse contains a Dog*
and demonstrates the four functions you will always need to define when your
class contains pointers: all necessary ordinary constructors, the
copy-constructor, operator= (either define it or disallow it), and a
destructor. The operator= checks for self-assignment as a matter of
course, even though it’s not strictly necessary here. This virtually
eliminates the possibility that you’ll forget to check for self-assignment
if you do change the code so that it matters.
In the example above, the
copy-constructor and operator= make a new copy of what the pointer points
to, and the destructor deletes it. However, if your object requires a lot of
memory or a high initialization overhead, you may want to avoid this copying. A
common approach to this problem is called reference
counting.
You give intelligence to the object that’s being pointed to so it knows
how many objects are pointing to it. Then copy-construction or assignment means
attaching another pointer to an existing object and incrementing the reference
count. Destruction means reducing the reference count and destroying the object
if the reference count goes to zero.
But what if you want to write to the
object (the Dog in the example above)? More than one object may be using
this Dog, so you’d be modifying someone else’s Dog as
well as yours, which doesn’t seem very neighborly. To solve this
“aliasing” problem, an additional technique
called copy-on-write is used. Before writing to a
block of memory, you make sure no one else is using it. If the reference count
is greater than one, you must make yourself a personal copy of that block before
writing it, so you don’t disturb someone else’s turf. Here’s a
simple example of reference counting and copy-on-write:
//: C12:ReferenceCounting.cpp
// Reference count, copy-on-write
#include "../require.h"
#include <string>
#include <iostream>
using namespace std;
class Dog {
string nm;
int refcount;
Dog(const string& name)
: nm(name), refcount(1) {
cout << "Creating Dog: " << *this << endl;
}
// Prevent assignment:
Dog& operator=(const Dog& rv);
public:
// Dogs can only be created on the heap:
static Dog* make(const string& name) {
return new Dog(name);
}
Dog(const Dog& d)
: nm(d.nm + " copy"), refcount(1) {
cout << "Dog copy-constructor: "
<< *this << endl;
}
~Dog() {
cout << "Deleting Dog: " << *this << endl;
}
void attach() {
++refcount;
cout << "Attached Dog: " << *this << endl;
}
void detach() {
require(refcount != 0);
cout << "Detaching Dog: " << *this << endl;
// Destroy object if no one is using it:
if(--refcount == 0) delete this;
}
// Conditionally copy this Dog.
// Call before modifying the Dog, assign
// resulting pointer to your Dog*.
Dog* unalias() {
cout << "Unaliasing Dog: " << *this << endl;
// Don't duplicate if not aliased:
if(refcount == 1) return this;
--refcount;
// Use copy-constructor to duplicate:
return new Dog(*this);
}
void rename(const string& newName) {
nm = newName;
cout << "Dog renamed to: " << *this << endl;
}
friend ostream&
operator<<(ostream& os, const Dog& d) {
return os << "[" << d.nm << "], rc = "
<< d.refcount;
}
};
class DogHouse {
Dog* p;
string houseName;
public:
DogHouse(Dog* dog, const string& house)
: p(dog), houseName(house) {
cout << "Created DogHouse: "<< *this << endl;
}
DogHouse(const DogHouse& dh)
: p(dh.p),
houseName("copy-constructed " +
dh.houseName) {
p->attach();
cout << "DogHouse copy-constructor: "
<< *this << endl;
}
DogHouse& operator=(const DogHouse& dh) {
// Check for self-assignment:
if(&dh != this) {
houseName = dh.houseName + " assigned";
// Clean up what you're using first:
p->detach();
p = dh.p; // Like copy-constructor
p->attach();
}
cout << "DogHouse operator= : "
<< *this << endl;
return *this;
}
// Decrement refcount, conditionally destroy
~DogHouse() {
cout << "DogHouse destructor: "
<< *this << endl;
p->detach();
}
void renameHouse(const string& newName) {
houseName = newName;
}
void unalias() { p = p->unalias(); }
// Copy-on-write. Anytime you modify the
// contents of the pointer you must
// first unalias it:
void renameDog(const string& newName) {
unalias();
p->rename(newName);
}
// ... or when you allow someone else access:
Dog* getDog() {
unalias();
return p;
}
friend ostream&
operator<<(ostream& os, const DogHouse& dh) {
return os << "[" << dh.houseName
<< "] contains " << *dh.p;
}
};
int main() {
DogHouse
fidos(Dog::make("Fido"), "FidoHouse"),
spots(Dog::make("Spot"), "SpotHouse");
cout << "Entering copy-construction" << endl;
DogHouse bobs(fidos);
cout << "After copy-constructing bobs" << endl;
cout << "fidos:" << fidos << endl;
cout << "spots:" << spots << endl;
cout << "bobs:" << bobs << endl;
cout << "Entering spots = fidos" << endl;
spots = fidos;
cout << "After spots = fidos" << endl;
cout << "spots:" << spots << endl;
cout << "Entering self-assignment" << endl;
bobs = bobs;
cout << "After self-assignment" << endl;
cout << "bobs:" << bobs << endl;
// Comment out the following lines:
cout << "Entering rename(\"Bob\")" << endl;
bobs.getDog()->rename("Bob");
cout << "After rename(\"Bob\")" << endl;
} ///:~The class Dog is the object
pointed to by a DogHouse. It contains a reference count and functions to
control and read the reference count. There’s a copy-constructor so you
can make a new Dog from an existing one.
The attach( ) function
increments the reference count of a Dog to indicate there’s another
object using it. detach( ) decrements the reference count. If the
reference count goes to zero, then no one is using it anymore, so the member
function destroys its own object by saying delete this.
Before you make any modifications (such
as renaming a Dog), you should ensure that you aren’t changing a
Dog that some other object is using. You do this by calling
DogHouse::unalias( ), which in turn calls
Dog::unalias( ). The latter function will return the existing
Dog pointer if the reference count is one (meaning no one else is
pointing to that Dog), but will duplicate the Dog if the reference
count is more than one.
The copy-constructor, instead of creating
its own memory, assigns Dog to the Dog of the source object. Then,
because there’s now an additional object using that block of memory, it
increments the reference count by calling
Dog::attach( ).
The operator= deals with an object
that has already been created on the left side of the =, so it must first
clean that up by calling detach( ) for that Dog, which will
destroy the old Dog if no one else is using it. Then operator=
repeats the behavior of the copy-constructor. Notice that it first checks to
detect whether you’re assigning the same object to
itself.
The destructor calls
detach( ) to conditionally destroy the Dog.
To implement copy-on-write, you must
control all the actions that write to your block of memory. For example, the
renameDog( ) member function allows you to change the values in the
block of memory. But first, it uses unalias( ) to prevent the
modification of an aliased Dog (a Dog with more than one
DogHouse object pointing to it). And if you need to produce a pointer to
a Dog from within a DogHouse, you unalias( ) that
pointer first.
main( ) tests the various
functions that must work correctly to implement reference counting: the
constructor, copy-constructor, operator=, and destructor. It also tests
the copy-on-write by calling renameDog( ).
Here’s the output (after a little
reformatting):
Creating Dog: [Fido], rc = 1
Created DogHouse: [FidoHouse]
contains [Fido], rc = 1
Creating Dog: [Spot], rc = 1
Created DogHouse: [SpotHouse]
contains [Spot], rc = 1
Entering copy-construction
Attached Dog: [Fido], rc = 2
DogHouse copy-constructor:
[copy-constructed FidoHouse]
contains [Fido], rc = 2
After copy-constructing bobs
fidos:[FidoHouse] contains [Fido], rc = 2
spots:[SpotHouse] contains [Spot], rc = 1
bobs:[copy-constructed FidoHouse]
contains [Fido], rc = 2
Entering spots = fidos
Detaching Dog: [Spot], rc = 1
Deleting Dog: [Spot], rc = 0
Attached Dog: [Fido], rc = 3
DogHouse operator= : [FidoHouse assigned]
contains [Fido], rc = 3
After spots = fidos
spots:[FidoHouse assigned] contains [Fido],rc = 3
Entering self-assignment
DogHouse operator= : [copy-constructed FidoHouse]
contains [Fido], rc = 3
After self-assignment
bobs:[copy-constructed FidoHouse]
contains [Fido], rc = 3
Entering rename("Bob")
After rename("Bob")
DogHouse destructor: [copy-constructed FidoHouse]
contains [Fido], rc = 3
Detaching Dog: [Fido], rc = 3
DogHouse destructor: [FidoHouse assigned]
contains [Fido], rc = 2
Detaching Dog: [Fido], rc = 2
DogHouse destructor: [FidoHouse]
contains [Fido], rc = 1
Detaching Dog: [Fido], rc = 1
Deleting Dog: [Fido], rc = 0By studying the output, tracing through
the source code, and experimenting with the program, you’ll deepen your
understanding of these techniques.
Because assigning an object to another
object of the same type is an activity most people expect to be possible,
the compiler will automatically create a type::operator=(type) if you
don’t make one. The behavior of this operator mimics that of the
automatically created copy-constructor; if the class contains objects (or is
inherited from another class), the operator= for those objects is called
recursively. This is called memberwise
assignment. For
example,
//: C12:AutomaticOperatorEquals.cpp
#include <iostream>
using namespace std;
class Cargo {
public:
Cargo& operator=(const Cargo&) {
cout << "inside Cargo::operator=()" << endl;
return *this;
}
};
class Truck {
Cargo b;
};
int main() {
Truck a, b;
a = b; // Prints: "inside Cargo::operator=()"
} ///:~The automatically generated
operator= for Truck calls
Cargo::operator=.
In general, you don’t want to let
the compiler do this for you. With classes of any sophistication (especially if
they contain pointers!) you want to explicitly create an operator=. If
you really don’t want people to perform assignment, declare
operator= as a
private
function. (You don’t need to define it unless you’re using it inside
the
class.)
In C and C++, if the compiler sees an
expression or function call using a type that isn’t quite the one it
needs, it can often perform an automatic type conversion from the type it has to
the type it
wants.
In C++, you can achieve this same effect for user-defined types by defining
automatic type conversion functions. These functions come in two flavors: a
particular type of constructor and an overloaded
operator.
If you define a
constructor that takes as its single argument an object
(or reference) of another type, that constructor allows the compiler to perform
an automatic type conversion. For example,
//: C12:AutomaticTypeConversion.cpp
// Type conversion constructor
class One {
public:
One() {}
};
class Two {
public:
Two(const One&) {}
};
void f(Two) {}
int main() {
One one;
f(one); // Wants a Two, has a One
} ///:~When the compiler sees f( )
called with a One object, it looks at the declaration for
f( ) and notices it wants a Two. Then it looks to see if
there’s any way to get a Two from a One, and it finds the
constructor Two::Two(One), which it quietly calls. The resulting
Two object is handed to f( ).
In this case, automatic type conversion
has saved you from the trouble of defining two overloaded versions of
f( ). However, the cost is the hidden constructor call to
Two, which may matter if you’re concerned about the efficiency of
calls to f( ).
There are times when automatic type
conversion via the constructor can cause problems. To turn it off, you modify
the constructor by prefacing with the keyword
explicit (which only works with constructors).
Used to modify the constructor of class Two in the example
above:
//: C12:ExplicitKeyword.cpp
// Using the "explicit" keyword
class One {
public:
One() {}
};
class Two {
public:
explicit Two(const One&) {}
};
void f(Two) {}
int main() {
One one;
//! f(one); // No auto conversion allowed
f(Two(one)); // OK -- user performs conversion
} ///:~By making Two’s constructor
explicit, the compiler is told not to perform any automatic conversion using
that particular constructor (other non-explicit constructors in that
class can still perform automatic conversions). If the user wants to make the
conversion happen, the code must be written out. In the code above,
f(Two(one)) creates a
temporary object of type
Two from one, just like the compiler did in the previous
version.
The second way to produce automatic type
conversion is through operator
overloading.
You can create a member function that takes the current type and converts it to
the desired type using the operator keyword followed by the type you want
to convert to. This form of operator overloading is unique because you
don’t appear to specify a return type – the return type is the
name of the operator you’re overloading. Here’s an
example:
//: C12:OperatorOverloadingConversion.cpp
class Three {
int i;
public:
Three(int ii = 0, int = 0) : i(ii) {}
};
class Four {
int x;
public:
Four(int xx) : x(xx) {}
operator Three() const { return Three(x); }
};
void g(Three) {}
int main() {
Four four(1);
g(four);
g(1); // Calls Three(1,0)
} ///:~With the constructor technique, the
destination class is performing the conversion, but with operators, the source
class performs the conversion. The value of the constructor technique is that
you can add a new conversion path to an existing system as you’re creating
a new class. However, creating a single-argument constructor always
defines an automatic type conversion (even if it’s got more than one
argument, if the rest of the arguments are defaulted), which may not be what you
want (in which case you can turn it off using explicit). In addition,
there’s no way to use a constructor conversion from a user-defined type to
a built-in type; this is possible only with operator
overloading.
One of the most convenient reasons to use
global overloaded operators instead of member operators
is that in the global versions, automatic type
conversion may be applied to either operand, whereas with member objects, the
left-hand operand must already be the proper type. If you want both operands to
be converted, the global versions can save a lot of coding. Here’s a small
example:
//: C12:ReflexivityInOverloading.cpp
class Number {
int i;
public:
Number(int ii = 0) : i(ii) {}
const Number
operator+(const Number& n) const {
return Number(i + n.i);
}
friend const Number
operator-(const Number&, const Number&);
};
const Number
operator-(const Number& n1,
const Number& n2) {
return Number(n1.i - n2.i);
}
int main() {
Number a(47), b(11);
a + b; // OK
a + 1; // 2nd arg converted to Number
//! 1 + a; // Wrong! 1st arg not of type Number
a - b; // OK
a - 1; // 2nd arg converted to Number
1 - a; // 1st arg converted to Number
} ///:~Class Number has both a member
operator+ and a friend operator–. Because
there’s a constructor that takes a single int argument, an
int can be automatically converted to a Number, but only under the
right conditions. In main( ), you can see that adding a
Number to another Number works fine because it’s an exact
match to the overloaded operator. Also, when the compiler sees a Number
followed by a + and an int, it can match to the member function
Number::operator+ and convert the int argument to a Number
using the constructor. But when it sees an int, a +, and a
Number, it doesn’t know what to do because all it has is
Number::operator+, which requires that the left operand already be a
Number object. Thus, the compiler issues an error.
With the friend
operator–, things are different. The compiler needs to fill in both
its arguments however it can; it isn’t restricted to having a
Number as the left-hand argument. Thus, if it sees
1 – a
it can convert the first argument to a
Number using the constructor.
Sometimes you want to be able to restrict
the use of your operators by making them members. For example, when multiplying
a matrix by a vector, the vector must go on the right. But if you want your
operators to be able to convert either argument, make the operator a friend
function.
Fortunately, the compiler will not take
1 – 1 and convert both arguments to Number objects and then
call operator–. That would mean that existing C code might suddenly
start to work differently. The compiler matches the “simplest”
possibility first, which is the built-in operator for the expression 1
–
1.
An example in which automatic type
conversion is extremely helpful occurs with any class that encapsulates
character strings (in this case, we will just implement the class using the
Standard C++ string class because it’s simple). Without automatic
type conversion, if you want to use all the existing string functions from the
Standard C library, you have to create a member function for each one, like
this:
//: C12:Strings1.cpp
// No auto type conversion
#include "../require.h"
#include <cstring>
#include <cstdlib>
#include <string>
using namespace std;
class Stringc {
string s;
public:
Stringc(const string& str = "") : s(str) {}
int strcmp(const Stringc& S) const {
return ::strcmp(s.c_str(), S.s.c_str());
}
// ... etc., for every function in string.h
};
int main() {
Stringc s1("hello"), s2("there");
s1.strcmp(s2);
} ///:~Here, only the strcmp( )
function is created, but you’d have to create a corresponding function for
every one in <cstring> that might be needed. Fortunately, you can
provide an automatic type conversion allowing access to all the functions in
<cstring>:
//: C12:Strings2.cpp
// With auto type conversion
#include "../require.h"
#include <cstring>
#include <cstdlib>
#include <string>
using namespace std;
class Stringc {
string s;
public:
Stringc(const string& str = "") : s(str) {}
operator const char*() const {
return s.c_str();
}
};
int main() {
Stringc s1("hello"), s2("there");
strcmp(s1, s2); // Standard C function
strspn(s1, s2); // Any string function!
} ///:~Now any function that takes a
char* argument can also take a Stringc argument because the
compiler knows how to make a char* from a
Stringc.
Because the compiler must choose how to
quietly perform a type conversion, it can get into trouble if you don’t
design your conversions correctly. A simple and obvious situation occurs with a
class X that can convert itself to an object of class Y with an
operator Y( ). If class Y has a constructor that takes a
single argument of type X, this represents the identical type conversion.
The compiler now has two ways to go from X to Y, so it will
generate an ambiguity error when that conversion
occurs:
//: C12:TypeConversionAmbiguity.cpp
class Orange; // Class declaration
class Apple {
public:
operator Orange() const; // Convert Apple to Orange
};
class Orange {
public:
Orange(Apple); // Convert Apple to Orange
};
void f(Orange) {}
int main() {
Apple a;
//! f(a); // Error: ambiguous conversion
} ///:~The obvious solution to this problem is
not to do it. Just provide a single path for automatic conversion from one type
to another.
A more difficult problem to spot occurs
when you provide automatic conversion to more than one type. This is sometimes
called
fan-out:
//: C12:TypeConversionFanout.cpp
class Orange {};
class Pear {};
class Apple {
public:
operator Orange() const;
operator Pear() const;
};
// Overloaded eat():
void eat(Orange);
void eat(Pear);
int main() {
Apple c;
//! eat(c);
// Error: Apple -> Orange or Apple -> Pear ???
} ///:~Class Apple has automatic
conversions to both Orange and Pear. The insidious thing about
this is that there’s no problem until someone innocently comes along and
creates two overloaded versions of eat( ). (With only one version,
the code in main( ) works fine.)
Again, the solution – and the
general watchword with automatic type conversion – is to provide only a
single automatic conversion from one type to another. You can have conversions
to other types; they just shouldn’t be automatic. You can create
explicit function calls with names like makeA( ) and
makeB( ).
Automatic type conversion can introduce
more underlying activities than you may expect. As a little brain teaser, look
at this modification of CopyingVsInitialization.cpp:
//: C12:CopyingVsInitialization2.cpp
class Fi {};
class Fee {
public:
Fee(int) {}
Fee(const Fi&) {}
};
class Fo {
int i;
public:
Fo(int x = 0) : i(x) {}
operator Fee() const { return Fee(i); }
};
int main() {
Fo fo;
Fee fee = fo;
} ///:~There is no constructor to create the
Fee fee from a Fo object. However, Fo has an automatic type
conversion to a Fee. There’s no copy-constructor to create a
Fee from a Fee, but this is one of the special functions the
compiler can create for you. (The default constructor, copy-constructor,
operator=, and destructor can be synthesized automatically by the
compiler.) So for the relatively innocuous statement
Fee fee = fo;
the automatic type conversion operator is
called, and a copy-constructor is created.
Use automatic type conversion carefully.
As with all operator overloading, it’s excellent when it significantly
reduces a coding task, but it’s usually not worth using
gratuitously.
The whole reason for the existence of
operator overloading is for those situations when it makes life easier.
There’s nothing particularly magical about it; the overloaded operators
are just functions with funny names, and the function calls happen to be made
for you by the compiler when it spots the right pattern. But if operator
overloading doesn’t provide a significant benefit to you (the creator of
the class) or the user of the class, don’t confuse the issue by adding
it.
Solutions to selected exercises
can be found in the electronic document The Thinking in C++ Annotated
Solution Guide, available for a small fee from www.BruceEckel.com.
[49]
Rob Murray, C++ Strategies & Tactics, Addison-Wesley, 1993, page
47.