为什么 SFINAE 不适用于此? [英] Why does SFINAE not apply to this?

问题描述

我在尝试 Visual Studio 10 (Beta 2) 时正在编写一些简单的点代码，并且我已经在我希望 SFINAE 启动的地方找到了这段代码，但似乎没有:

I'm writing some simple point code while trying out Visual Studio 10 (Beta 2), and I've hit this code where I would expect SFINAE to kick in, but it seems not to:

template<typename T>
struct point {
    T x, y;
    point(T x, T y) : x(x), y(y) {}
};

template<typename T, typename U>
struct op_div {
    typedef decltype(T() / U()) type;
};

template<typename T, typename U>
point<typename op_div<T, U>::type>
operator/(point<T> const& l, point<U> const& r) {
    return point<typename op_div<T, U>::type>(l.x / r.x, l.y / r.y);
}

template<typename T, typename U>
point<typename op_div<T, U>::type>
operator/(point<T> const& l, U const& r) {
    return point<typename op_div<T, U>::type>(l.x / r, l.y / r);
}

int main() {
    point<int>(0, 1) / point<float>(2, 3);
}

这会导致 error C2512: 'point<T>::point' : 没有合适的默认构造函数可用

鉴于它是一个测试版，我用在线的 Comeau 编译器做了一个快速的健全性检查，它同意一个相同的错误，所以这种行为似乎是正确的，但我不明白为什么.

Given that it is a beta, I did a quick sanity check with the online comeau compiler, and it agrees with an identical error, so it seems this behavior is correct, but I can't see why.

在这种情况下，一些解决方法是简单地内联 decltype(T()/U())，给点类一个默认构造函数，或者在完整的结果表达式上使用 decltype，但是我在尝试简化不需要默认构造函数 * 的 op_div 版本时遇到的错误时遇到了这个错误，所以我宁愿修正我对 C++ 的理解，而不是只做有效的事情.

In this case some workarounds are to simply inline the decltype(T() / U()), to give the point class a default constructor, or to use decltype on the full result expression, but I got this error while trying to simplify an error I was getting with a version of op_div that did not require a default constructor*, so I would rather fix my understanding of C++ rather than to just do what works.

谢谢！

*:原文:

template<typename T, typename U>
struct op_div {
    static T t(); static U u();
    typedef decltype(t() / u()) type;
};

这会导致 错误 C2784:'point::type>运算符/(const point<T>,const U &)':无法推导出 'const point<T> 的模板参数.&'来自'int'，也用于point<T>/point 重载.

Which gives error C2784: 'point<op_div<T,U>::type> operator /(const point<T> &,const U &)' : could not deduce template argument for 'const point<T> &' from 'int', and also for the point<T> / point<U> overload.

推荐答案

不是 100% 确定.似乎编译器需要实例化两个重载以确定哪个更好，但是在尝试使用 T = int 和 U = point 实例化另一个 op_div 时，这会导致 SFINAE 未涵盖的错误(在这种情况下，错误不是 op_div 没有类型，而是无法确定该类型).

Not 100% sure. It appears that the compiler needs to instantiate both overloads to determine which is better, but while trying to instantiate the other op_div with T = int and U = point<float>, this leads to an error that is not covered by SFINAE (the error is not that op_div doesn't have type in this case, but that type cannot be determined).

如果第二种类型是点 (boost::disable_if)，您可以尝试禁用第二个重载.

You could try to disable the second overload if the second type is a point (boost::disable_if).

此外，似乎可行的是推迟返回类型声明(取消 op_div 结构，但取决于您的编译器支持哪些 C++0x 功能):

Also, what seems to work is postponed return type declaration (doing away with the op_div struct, but depending on which C++0x features are supported by your compiler):

template<typename T, typename U>
auto
operator/(point<T> const& l, point<U> const& r) -> point<decltype(l.x / r.x)> {
    return {l.x / r.x, l.y / r.y};
}

template<typename T, typename U>
auto
operator/(point<T> const& l, U const& r) -> point<decltype(l.x / r)> {
    return {l.x / r, l.y / r};
}

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