Category : C++ Source Code
Archive   : C_ALL.ZIP
Filename : TI638.ASC

PRODUCT : C++ NUMBER : 638
VERSION : All
OS : PC DOS
DATE : August 12, 1992 PAGE : 1/2

TITLE : Calling Operators from Member Functions




The following provides examples of different ways to invoke an
operator, how to call such operators from within a member
function and deals with some related multiple inheritance issues.

You can always invoke an operator by name. For example, given:

X_Class {
...
public:
X_Class &operator +( X_Class y);
};

X_Class x, y;

the function for the + operator can be invoked in several ways:

x + y;
x.operator +(y);

We can think of the second example in this way: not only have we
overloaded the + operator to add together instances of the
X_Class, we also created a member function of the X_Class whose
name is operator +.

If X_Class has a member function foo, and we want to call the +
operator from within it, we would do it by using its member
function name:

X_Class::foo(X_Class &y) {
...
operator +(y);
...
}

If one is dealing with multiple inheritance, and operator + has
been redefined in several different classes, one might get an
ambiguity at some point. To resolve this, or any other ambiguity
the compiler faces when it cannot determine which of the
overloaded functions it must use, you can qualify a member
function name with its class name. For example suppose we have:
















PRODUCT : C++ NUMBER : 638
VERSION : All
OS : PC DOS
DATE : August 12, 1992 PAGE : 2/2

TITLE : Calling Operators from Member Functions




A B
\ /
C

and there is a public A & operator +( int ) defined for both A
and B, and both A and B are inherited publicly into C. Then

A ans;
C c1;

ans = c1 + 4;

will be ambiguous. It can be resolved by going,

ans = c1.B::operator +(4); or

ans = c1.A::operator +(4);

depending on which one you wanted to use. This is because :: is
the scope operator. It is used to specify which scope a given
symbol comes from.

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