c++ - 为什么 std::add_lvalue_reference 的行为不如预期？

Question

#include <type_traits>

template<typename T>
using Ref1 = T & ;

template<typename T>
using Ref2 = std::add_lvalue_reference_t<T>;

template<typename T>
void f1(Ref1<T>)
{}

template<typename T>
void f2(Ref2<T>)
{}

int main()
{
    int n{};
    f1(n); // ok
    f2(n); // error
}

我的编译器是clang 7.0，用c++17编译。错误信息是:

error : no matching function for call to 'f2'
  note: candidate template ignored:
        couldn't infer template argument 'T'

为什么 f1 可以，但 f2 不行？

Answer 1

std::add_lvalue_reference_t<T> 定义为 typename std::add_lvalue_reference<T>::type , 然后为 template<typename T> void f2(Ref2<T>) ，即 template<typename T> void f2(typename std::add_lvalue_reference<T>::type) , 属于 non-deduced context ，这会导致模板参数推导失败。

In the following cases, the types, templates, and non-type values that are used to compose P do not participate in template argument deduction, but instead use the template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

1) The nested-name-specifier (everything to the left of the scope resolution operator ::) of a type that was specified using a qualified-id: