同时定义运算符void *和运算符bool [英] define both operator void* and operator bool

问题描述

我尝试用一​​个operator bool和一个operator void*创建一个类，但是编译器说它们是模棱两可的.有什么方法可以向编译器解释要使用的运算符，还是不能同时使用它们?

I tried creating a class with one operator bool and one operator void*, but the compiler says they are ambigous. Is there some way I can explain to the compiler what operator to use or can I not have them both?

class A {
public:
    operator void*(){
        cout << "operator void* is called" << endl;
        return 0;
    }

    operator bool(){
        cout << "operator bool is called" << endl;
        return true;
    }
};

int main()
{
    A a1, a2;
    if (a1 == a2){
        cout << "hello";
    }
} 

推荐答案

您可以直接致电运营商.

You could call the operator directly.

int main()
{
    A a1, a2;
    if (static_cast<bool>(a1) == static_cast<bool>(a2)){
        cout << "hello";
    }
} 

但是，在这种情况下，您似乎应该定义operator==()而不是依赖于转换.

In this case, though, it looks like you should define operator==() and not depend on conversions.

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