模板constexpr推导

Question

在下面的代码中，为何a中的fn4处于fn1，fn2和fn3中时不被视为编译时常量？ 编译为clang++ -std=c++17 file.cxx

template<typename T>      constexpr T fn1(T a) {return a;}
template<typename T, T a> constexpr T fn2()    {return a;}
template<typename T>      constexpr T fn3(T a) {return fn1<T>(a);}
template<typename T>      constexpr T fn4(T a) {return fn2<T, a>();}

int main() {
    auto constexpr a = fn1<int>(1);
    auto constexpr b = fn2<int, 2>();
    auto constexpr c = fn3<int>(3);
    auto constexpr d = fn4<int>(4);
    return 0;
}

使用clang-7.0输出：

file.cxx:9:56: error: no matching function for call to 'fn2'
template<typename T>      constexpr T fn4(T a) {return fn2<T, a>();}
                                                       ^~~~~~~~~
file.cxx:15:21: note: in instantiation of function template specialization 'fn4<int>' requested here
        auto constexpr d = fn4<int>(4);
                           ^
file.cxx:7:39: note: candidate template ignored: invalid explicitly-specified argument for template parameter 'a'
template<typename T, T a> constexpr T fn2()    {return a;}
                                      ^
file.cxx:15:17: error: constexpr variable 'd' must be initialized by a constant expression
        auto constexpr d = fn4<int>(4);
                       ^   ~~~~~~~~~~~
2 errors generated.