C++20概念构造函数重载

Question

让

#include<concepts>

struct B{};

template<typename T>
struct A{
  A(int) {}
};

template<>
struct A<B>{
  A(B d) {}
};

template<typename T>
struct C{
    A<T> a;
    C(B b) requires std::same_as<T, B> : a{b} {}
    C(int i) requires (!std::same_as<T, B>) : a{i} {}
};

template class C<int>;

int main() {}

当我尝试使用 gcc v.10.2 编译它时，出现以下错误：

main.cpp: In instantiation of ‘C<T>::C(B) requires  same_as<T, B> [with T = int]’:
main.cpp:22:16:   required from here
main.cpp:18:45: error: no matching function for call to ‘A<int>::A(<brace-enclosed initializer list>)’
   18 |     C(B b) requires std::same_as<T, B> : a{b} {}
      |                                             ^
main.cpp:7:3: note: candidate: ‘A<T>::A(int) [with T = int]’
    7 |   A(int) {}
      |   ^
main.cpp:7:5: note:   no known conversion for argument 1 from ‘B’ to ‘int’
    7 |   A(int) {}
      |     ^~~
main.cpp:6:8: note: candidate: ‘constexpr A<int>::A(const A<int>&)’
    6 | struct A{
      |        ^
main.cpp:6:8: note:   no known conversion for argument 1 from ‘B’ to ‘const A<int>&’
main.cpp:6:8: note: candidate: ‘constexpr A<int>::A(A<int>&&)’
main.cpp:6:8: note:   no known conversion for argument 1 from ‘B’ to ‘A<int>&&’

但 C++20 中的概念不会阻止实例化不满足其约束的函数？。我知道 SFINAE 不起作用，因为这个构造函数不是模板，但概念应该也适用于普通函数。

我做错了什么？

Answer 1

好的，这是 gcc 10.2 的一个错误，我需要至少更新到 10.3

C++20概念构造函数重载

c++20 concepts constructor overloading

c++

gcc

c++20

c++-concepts

gcc10