c++ - 替换失败是依赖非类型模板参数的错误吗？

Question

假设我有这些模板别名:

enum class enabler {};

template <typename T>
using EnableIf = typename std::enable_if<T::value, enabler>::type;
template <typename T>
using DisableIf = typename std::enable_if<!T::value, enabler>::type;

我可以在 GCC 中执行以下操作:

#include <iostream>

template <typename T, EnableIf<std::is_polymorphic<T>> = {}>
void f(T) { std::cout << "is polymorphic\n"; }

template <typename T, DisableIf<std::is_polymorphic<T>> = {}>
void f(T) { std::cout << "is not polymorphic\n"; }

struct foo { virtual void g() {} };

int main() {
    f(foo {});
    f(int {});
}

它打印:

is polymorphic
is not polymorphic

这符合我的期望。

使用 clang 时代码无法编译。它会产生以下错误消息。

test.cpp:11:58: error: expected expression
template <typename T, EnableIf<std::is_polymorphic<T>> = {}>
                                                         ^
test.cpp:14:59: error: expected expression
template <typename T, DisableIf<std::is_polymorphic<T>> = {}>
                                                          ^
test.cpp:20:3: error: no matching function for call to 'f'
  f(foo {});
  ^
test.cpp:12:6: note: candidate template ignored: couldn't infer template argument ''
void f(T) { std::cout << "is polymorphic\n"; }
     ^
test.cpp:15:6: note: candidate template ignored: couldn't infer template argument ''
void f(T) { std::cout << "is not polymorphic\n"; }
     ^
test.cpp:21:3: error: no matching function for call to 'f'
  f(int {});
  ^
test.cpp:12:6: note: candidate template ignored: couldn't infer template argument ''
void f(T) { std::cout << "is polymorphic\n"; }
     ^
test.cpp:15:6: note: candidate template ignored: couldn't infer template argument ''
void f(T) { std::cout << "is not polymorphic\n"; }
     ^
4 errors generated.

它应该编译吗？这两个编译器哪个有问题？

Answer 1

^{首先，感谢@ Richard Smith在 #llvm IRC Channel on oftc 上的解释。}
不幸的是，这不是合法的 C++，因此 Clang 是正确的:{} 不是表达式而是 braced-init-list 因此永远不会是 constant 非类型模板参数的初始化程序中需要的表达式。

§14.3.2 [temp.arg.non-type] p1

A template-argument for a non-type, non-template template-parameter shall be one of:

• for a non-type template-parameter of integral or enumeration type, a converted constant expression (5.19) of the type of the template-parameter; or
• [...]

一个解决方案是在 enabler 中添加一个虚拟值。