enable_if g++ 和 clang++ 之间的区别

Question

我想编写一个 returns 类型 T 实例的函数，但根据 T 的构造方式表现不同。假设我有这样的结构

#include <type_traits>                                                                                                                                                                                                                                                                                                        
#include <iostream>

struct A {}; 
struct B {}; 
struct C { 
  C(A a) {
    std::cout << "C" << std::endl;
  }
};

我想通过给它们一个 A 来创建 C。我有一个这样的结构，它使用 enable_if 选择两个函数之一：

struct E {

  template< bool condition = std::is_constructible<C, A>::value,std::enable_if_t<condition,int> = 0>
  C get() {
    return C{A{}};
  }
  template< bool condition = std::is_constructible<C, B>::value,std::enable_if_t<condition,bool> = false>
  C get() {
    return C{B{}};
  }
};

这可以用 g++82（我认为也是 g++9）编译得很好，但是 clang9 给我错误

$ clang++ --std=c++17 main.cpp 
main.cpp:26:12: error: no matching constructor for initialization of 'C'
    return C{B{}};
           ^~~~~~
main.cpp:6:8: note: candidate constructor (the implicit copy constructor) not viable: no known conversion from 'B' to 'const C' for 1st argument
struct C {
       ^
main.cpp:6:8: note: candidate constructor (the implicit move constructor) not viable: no known conversion from 'B' to 'C' for 1st argument
struct C {
       ^
main.cpp:7:3: note: candidate constructor not viable: no known conversion from 'B' to 'A' for 1st argument
  C(A a) {
  ^
1 error generated.

即使 enable_if 应该隐藏该功能。（我叫E e; auto c = e.get();）。如果我不对 C 进行硬编码，而是使用模板传入 C，那么它在两个编译器中都可以工作。

template<typename T>
struct F {

  template< bool condition = std::is_constructible<T, A>::value,std::enable_if_t<condition,int> = 0>
  T get() {
    return T{A{}};
  }
  template< bool condition = std::is_constructible<T, B>::value,std::enable_if_t<condition,bool> = false>
  T get() {
    return T{B{}};
  }
};

我不明白为什么 clang 显然要对函数主体进行类型检查，即使该函数应该被 enable_if 禁用。

Answer 1

因为您似乎可以访问支持 c++17 的编译器，您可以使用 if constexpr 而不是 enable_if 来完成您想要的。

#include <iostream>
#include <type_traits>

struct A {};

struct B {};

struct C {
    explicit C(A a) {
        std::cout << "C" << std::endl;
    }
};

template<typename T>
struct False : std::false_type {};

struct E {
    template<typename T = void>
    C get() const {
        if constexpr (std::is_constructible_v<C, A>) {
            return C{A{}};
        } else if constexpr (std::is_constructible_v<C, B>) {
            return C{B{}};
        } else {
            static_assert(False<T>::value, "Error");
        }
    }
};

int main() {
    const auto C{E{}.get()};
    return 0;
}

Answer 2

两个编译器都对，

http://eel.is/c++draft/temp.res#8.1

The validity of a template may be checked prior to any instantiation. [ Note: Knowing which names are type names allows the syntax of every template to be checked in this way. — end note ] The program is ill-formed, no diagnostic required, if:

(8.1)
- no valid specialization can be generated for a template or a substatement of a constexpr if statement within a template and the template is not instantiated, or

[..]

(8.4)
- a hypothetical instantiation of a template immediately following its definition would be ill-formed due to a construct that does not depend on a template parameter, or

return C{B{}}; 不依赖于模板，而且是错误的。通过诊断问题，clang 很好。

enable_if g++ 和 clang++ 之间的区别

Difference between g++ and clang++ with enable_if

c++

clang++

enable-if