重载运算符 - > [英] Overloading operator ->

问题描述

这是我的代码示例：

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 '.'
}

为什么编译器试图 >从Y到X？当我实现X :: op->然后我不能返回X那里 - 编译错误说无限递归，而从X :: op->返回一些Z再次说，Z没有operator->，因此更高和

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.

任何人都可以解释这个有趣的行为吗？ ：）

Can anyone explain this interesting behavior? :)

推荐答案

问题是， operator - > 以返回指针，而不是 。这个想法是 operator - > 应该返回一个指向应该有指针的实际对象的指针。例如，对于具有重载运算符 - > 的类，代码

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;

可转换为

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

您的代码的问题是 operator - > 返回一个引用，所以写

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

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

完全合法，因为您明确调用该运算符，但写

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

myClass->f();

是非法的，因为编译器尝试将其扩展到

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

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

，返回类型 operator-> 不是指针。

为了解决这个问题，修改代码，以便在 operator - > 。如果你想重载一个操作符来返回一个引用，重载 operator * ;指针解引用确实会产生引用。

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.

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