SFINAE编译器故障 [英] SFINAE compiler troubles

问题描述

下面的代码应该检测 T 是否具有 begin 和 end 方法：

The following code of mine should detect whether T has begin and end methods:

template <typename T>
struct is_container
{
    template <typename U, typename U::const_iterator (U::*)() const,
                          typename U::const_iterator (U::*)() const>
    struct sfinae {};

    template <typename U> static char test(sfinae<U, &U::begin, &U::end>*);
    template <typename U> static long test(...);

    enum { value = (1 == sizeof test<T>(0)) };
};

这里是一些测试代码：

#include <iostream>
#include <vector>
#include <list>
#include <set>
#include <map>

int main()
{
    std::cout << is_container<std::vector<std::string> >::value << ' ';
    std::cout << is_container<std::list<std::string> >::value << ' ';
    std::cout << is_container<std::set<std::string> >::value << ' ';
    std::cout << is_container<std::map<std::string, std::string> >::value << '\n';
}

在g ++ 4.5.1上，输出为 1 1 1 1 。然而，在Visual Studio 2008上，输出为 1 1 0 0 。我做错了什么，或者这只是一个VS 2008错误？任何人可以测试不同的编译器？谢谢！

On g++ 4.5.1, the output is 1 1 1 1. On Visual Studio 2008, however, the output is 1 1 0 0. Did I do something wrong, or is this simply a VS 2008 bug? Can anyone test on a different compiler? Thanks!

推荐答案

Stephan T. Lavavej有 this to say：

Stephan T. Lavavej has this to say:

请注意，技术上禁止使用标准库成员函数的地址。（可以重载它们，使& foo :: bar 含糊不清，并且可以有其他默认参数，通过 static_cast 可以消除歧义。）

Please note that it is technically forbidden to take the address of a Standard Library member function. (They can be overloaded, making &foo::bar ambiguous, and they can have additional default arguments, defeating attempts to disambiguate via static_cast.)

所以我想使用更简单的版本，只检查嵌套的 const_iterator 类型。

So I guess I'm going to use the simpler version that only checks for the nested const_iterator type.

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