c++ - 引导构造函数的外定义(c++20)

Question

g++ 愉快地接受以下代码，而 clang 和 msvc 能够匹配行外定义。
知道为什么吗？

template <bool B>
struct test 
{
    test() requires (B);
    test() requires(!B);
};


template <>
test<true>::test()
{}

template <>
test<false>::test()
{}

int main()
{
    test<false> a;
    test<true> b;
    return 0;
}

Demo
铛:

error: out-of-line definition of 'test' does not match any declaration in 'test<true>'

msvc:

error C2244: 'test<true>::test': unable to match function definition to an existing declaration

Answer 1

您正在声明受约束的构造函数，但定义了两个不受约束的特化。那些永远不会匹配。
你的意思可能是:

template <bool B>
struct test
{
    test() requires (B);
    test() requires(!B);
};

template <bool B>
test<B>::test() requires (B)
{}

template <bool B>
test<B>::test() requires (!B)
{}

这在所有 3 个编译器中都可以正常编译。
至于为什么你的原始版本编译 - 这是一个 GCC 错误 96830 . Clang 是对的，代码格式错误，因为外部定义与模板定义不匹配(还要注意 template<> ... 是完整的特化语法)。
见 [temp.class.general]/3 (强调我的):

When a member of a class template is defined outside of the class template definition, the member definition is defined as a template definition with the template-head equivalent to that of the class template.

[temp.over.link]/6 :

Two template-heads are equivalent if their template-parameter-lists have the same length, corresponding template-parameters are equivalent and are both declared with type-constraints that are equivalent if either template-parameter is declared with a type-constraint, and if either template-head has a requires-clause, they both have requires-clauses and the corresponding constraint-expressions are equivalent.

另见 [temp.mem.func]/1例如，声明一个外联的受约束成员:

template<typename T> struct S {
    void f() requires C<T>;
    void g() requires C<T>;
};

template<typename T>
void S<T>::f() requires C<T> { }      // OK
template<typename T>
void S<T>::g() { }                    // error: no matching function in S<T>