c – 为什么成员函数temporaries不绑定到正确的类型？

发布时间：2020-12-16 04:58:15 所属栏目：百科来源：网络整理

导读：假设我们有以下基类和派生类： #include string#include iostreamclass Car {public: void Drive() { std::cout "Baby,can I drive your car?" std::endl; }};class Porsche : public Car {}; ..还有以下模板功能： template typename T,typename Vvoid Func

假设我们有以下基类和派生类：

#include <string>
#include <iostream>

class Car {
public:
    void Drive() { std::cout << "Baby,can I drive your car?" << std::endl; }
};

class Porsche : public Car {
};

..还有以下模板功能：

template <typename T,typename V>
void Function(void (T::*m1)(void),void (V::*m2)(void)) {
    std::cout << (m1 == m2) << std::endl;
}

为什么使用GCC编译：

int main(int argc,char** argv) {
    void (Porsche::*ptr)(void) = &Porsche::Drive;
    Function(ptr,ptr);
    return 0;
}

……但不是吗？

int main(int argc,char** argv) {
    void (Porsche::*ptr)(void) = &Porsche::Drive;
    Function(&Porsche::Drive,ptr);
    return 0;
}

解决方法

int main(int argc,ptr);
    return 0;
}

ptr的类型为void(Porsche :: *)(),但& Porsche :: Drive的类型为void(Car :: *)()(因为该成员在Car中找到,而不是在Porsche中).因此,调用的函数将这两个成员指针与这些类型进行比较,标准说

In addition,pointers to members can be compared,or a pointer to member and a null pointer constant. Pointer to member conversions (4.11) and qualification conversions (4.4) are performed to bring them to a common type. If one operand is a null pointer constant,the common type is the type of the other operand. Otherwise,the common type is a pointer to member type similar (4.4) to the type of one of the operands,with a cv-qualification signature (4.4) that is the union of the cv-qualification signatures of the operand types.

4.11描述了从void(Base :: *)()到void(Derived :: *)()的隐式标准转换.因此,比较会找到常见类型void(Porsche :: *)().对于Porsche类型的对象,两个成员指针都将引用相同的函数(即Car :: Drive) – 因此比较将产生true. comeau web compiler遵循此解释并编译您的代码.

（编辑：李大同）

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