Question

我正在尝试实现is_base模板，我有“一点问题”。为什么这不起作用呢？

#include <iostream>
using std::cout;
class Car
{
};

class Fiesta : public Car
{
};

template<class Base, class Derived>
struct isBase
{
    typedef char yes[2];
    typedef char no[1];

    template<class B, class D>
    static yes& f(D* d, B* b = d);

    template<class,class>
    static no&  f(...);

    static  bool type;
};




template<class Base, class Derived>
bool isBase<Base,Derived>::type = (sizeof(f<Base,Derived>(0,0)) == sizeof(yes));

int _tmain(int argc, _TCHAR* argv[])
{
    cout << isBase<Fiesta,Car>::type;//It should say false here but says true

    return 0;
}

Answer 1

您明确地为指针提供了一个值：f<Base,Derived>(0, 0)，此处不需要进行转换，尤其不是派生到基础的转换;此测试将始终通过，因为第一个测试始终是可调用的（任何指针都可以为null）。

你想要这样的东西：

template<class Base, class Derived>
struct isBase
{
    typedef char yes[2];
    typedef char no[1];

    template <class B>
    static yes& f(B*);

    template <class>
    static no&  f(...);

    // (1) make it const (as it should be) and just define it inline
    // (2) it's not a type, it's a value; name appropriately
    static const bool value = sizeof(f<Base>((Derived*)0)) == sizeof(yes);
};

// use a standard signature for main
int main() // if you don't need the arguments, don't list them
{
    cout << isBase<Fiesta, Car>::value;
    cout << isBase<Car, Fiesta>::value;

   // return 0 is implicit for main, helpful for snippets
}

如果指针类型是或可以转换为Base*，则将调用第一个重载。所以我们创建一个类型为Derived*的指针，如果它实际上是一个派生类，那么转换将起作用，并调用第一个重载;否则，第二个。

SFINAE构造问题

1 个答案: