Overloading operator ->

问题

Here is my code example:

class X
{
public:
        void f() {}
};

class Y : public X
{
public:
        X& operator->() { return *this; }
        void f() {}
};

int main()
{
        Y t;
        t.operator->().f(); // OK
        t->f(); // error C2819: type 'X' does not have an overloaded member 'operator ->'
                // error C2232: '->Y::f' : left operand has 'class' type, use '.'
}

Why the compiler is trying to "move the responsibility" for operator-> from Y to X? When I implement X::op-> then I cannot return X there - compile error says "infinite recursion" while returning some Z from X::op-> again says that Z doesn't have operator->, thus going higher and higher in hierarchy.

Can anyone explain this interesting behavior? :)

回答1:

The problem is that operator -> is supposed to return a pointer, not a reference. The idea is that operator -> should return a pointer to the real object that should have the pointer applied to it. For example, for a class with an overloaded operator ->, the code

myClass->myValue;

translates into

(myClass.operator-> ())->myValue;

The problem with your code is that operator -> returns a reference, so writing

myClass.operator->().f();

is perfectly legal because you're explicitly invoking the operator, but writing

myClass->f();

is illegal, because the compiler is trying to expand it to

myClass.operator->()->f();

and the return type of operator-> isn't a pointer.

To fix this, change your code so that you return a pointer in operator ->. If you want to overload an operator to return a reference, overload operator *; pointer dereferences should indeed produce references.

回答2:

Because that's how overloaded -> works in C++.

When you use overloaded ->, expression a->b is translated into a.operator->()->b. This means that your overloaded operator -> must return something that will itself support another application of operator ->. For this reason a single invocation of overloaded -> might turn into a long chain of invocations of overloaded ->s until it eventually reaches an application of built-in ->, which ends the chain.

In your case you need to return X* from your overloaded ->, not X&.

回答3:

The syntax is wrong, should be:

T->T2

T2* T::operator ->();

Look at wikipedia's article: Operators in C and C++

If you want to overload, you must use the right syntax for the overloaded operator

回答4:

You probably want:

class Y : public X
{
public:
        X* operator->() { return this; }
        void f() {}
};

来源：https://stackoverflow.com/questions/4896238/overloading-operator

标签

c++

overloading

operator-keyword