SFINAE误差

Question

为什么以下代码会导致错误？我认为编译器只是在这里选择适当的重载吗？

#include <iostream>
using std::cout;
using std::endl;

template <typename ToCheckFor>
struct InterfaceCheck {

    // used by the constexpr function, the function will pass in a pointer to
    // a type with the required types
    template <typename _ToCheckFor, void (_ToCheckFor::*) ()>
    struct InterfaceCheckImplTag {};

    // used to check for the presence of a function print()
    // template <typename T>
    // static constexpr bool function(__attribute__((unused)) void* ptr) {}
    template <typename T>
    static constexpr bool function(__attribute__((unused)) void* ptr) { 
        return false; 
    }
    template <typename T>
    static constexpr bool function (__attribute__((unused))  
         InterfaceCheckImplTag<T, &T::print>* ptr) { 

        return true; 
    }

    constexpr static const bool value = function<ToCheckFor>(nullptr);
};

struct Something {
    void print() { cout << "Something::print()" << endl; }
};

int main() {

    cout << InterfaceCheck<Something>::value << endl;

    return 0;
}

为什么用省略号替换void*参数使代码像预期的那样工作呢？因此，下面的代码按预期工作

#include <iostream>
using std::cout;
using std::endl;

template <typename ToCheckFor>
struct InterfaceCheck {

    // used by the constexpr function, the function will pass in a pointer to
    // a type with the required types
    template <typename _ToCheckFor, void (_ToCheckFor::*) ()>
    struct InterfaceCheckImplTag {};

    // used to check for the presence of a function print()
    // template <typename T>
    // static constexpr bool function(__attribute__((unused)) void* ptr) {}
    template <typename T>
    static constexpr bool function(...) { 
        return false; 
    }
    template <typename T>
    static constexpr bool function (__attribute__((unused))  
         InterfaceCheckImplTag<T, &T::print>* ptr) { 

        return true; 
    }

    constexpr static const bool value = function<ToCheckFor>(nullptr);
};

struct Something {
    void print() { cout << "Something::print()" << endl; }
};

int main() {

    cout << InterfaceCheck<Something>::value << endl;

    return 0;
}

Answer 1

为什么以下代码会导致错误？

对于function重载有两种可行的选择。这两种方法都涉及从所提供的参数进行转换，而且这两种转换都优于另一种：

error: call to 'function' is ambiguous
    constexpr static const bool value = function<ToCheckFor>(nullptr);
                                        ^~~~~~~~~~~~~~~~~~~~
test.cpp:36:13: note: in instantiation of template class 'InterfaceCheck<Something>' requested here
    cout << InterfaceCheck<Something>::value << endl;
            ^
test.cpp:17:27: note: candidate function [with T = Something]
    static constexpr bool function(__attribute__((unused)) void* ptr) { 
                          ^
test.cpp:21:27: note: candidate function [with T = Something]
    static constexpr bool function (__attribute__((unused))  

使用function(...)进行修复是有效的，因为从任何东西到...的转换总是比其他任何东西都“更糟糕”(但仍然合法)。一旦你知道了这是个绝妙的把戏。

从13.3.3.2排序隐式转换序列over.ics.rank

1. 当比较隐式转换序列的基本形式时(如13.3.3.1所定义)

- a standard conversion sequence (13.3.3.1.1) is a better conversion sequence than a user-defined conversion sequence or an ellipsis   conversion sequence, and
- a user-defined conversion sequence (13.3.3.1.2) is a better conversion sequence than an ellipsis conversion sequence   (13.3.3.1.3).

​

历史

我第一次从现代C++设计第2.7节学到了这个技巧。我不知道这是否是发明它的地方。但这也不是个糟糕的猜测。这本书现在已有15年历史了，它仍然是一本很好的读物。