为什么 sizeof... 不能使用这个别名模板？

Question

我想将可变参数函数限制为一定数量的输入 - 比如两个。为此，这在我的环境中运行良好（VS2017，C++17）：

#include <type_traits>

template<typename... T> 
auto f(T...) -> typename std::enable_if<sizeof...(T) == 2>::type {
    // no-op
}

int main() {
    // f(1); // should fail
    f(1,2);
    // f(1,2,3); // should fail
}

但是如果我引入别名模板，它不会。

#include <type_traits>

template<typename... T> 
using two_params = typename std::enable_if<sizeof...(T) == 2>::type; 


template<typename... T> 
auto f(T...) -> two_params<T...> { // failed to specialize alias template

}

int main() {
    // f(1); // should fail
    f(1,2);
    // f(1,2,3); // should fail
}

有趣的是，如果我将条件更改为1 或实际需要的输入数量，则替换成功。

// This works, except that it permits a single argument even when it shouldn't.
// Both conditions ||'d together seems to be needed in the general case.
template<typename... T> 
using two_params = typename std::enable_if<sizeof...(T) == 1 || sizeof...(T) == 2>::type; 

好像f(1,2)生成了两个sizeof...(T)的值。这到底是怎么回事？

我看过的一些参考文献：

• sizeof...
• alias template
• parameter pack
• （类似问题）

Answer 1

Microsoft 的 Jonathan Emmett 刚刚 confirmed 说这是一个编译器错误：

Thanks for this report. I can confirm that this is a compiler bug with alias templates and pack expansions. We are currently working on a major set of fixes for alias specializations, and this work is currently slated to be included in the VS 2017 15.9 release. I can also confirm that this bug is fixed as part of this rework.