c++ - "requires"忽略一个字段不是静态的

Question

考虑以下代码:

#include <iostream>

constexpr int fun(int const&) { return 5; }
struct T { int x; };

int main() {
  std::cout << fun(T::x) << std::endl;                   // Line A
  std::cout << requires { fun(T::x); } << std::endl;     // Line B
}

如果我只注释 B 行，则无法编译代码(正如预期的那样，因为 x 在 T 中不是静态的)。但是，当我只注释 A 行时，代码在 Clang 11.0.0 和 GCC 10.2.0(均输出 1)下编译得很好。我想念的是什么 requires ?它不应该返回false ?

Answer 1

一个 requires 表达式是一大堆未计算的操作数。

[expr.prim.req]

2 A requires-expression is a prvalue of type bool whose value is described below. Expressions appearing within a requirement-body are unevaluated operands.

命名非静态数据成员的限定 ID 总是可以出现在未计算的操作数中。

[expr.prim.id]

2 An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member's class or a class derived from that class, or

• to form a pointer to member ([expr.unary.op]), or

• if that id-expression denotes a non-static data member and it appears in an unevaluated operand. [ Example:

struct S {
  int m;
};
int i = sizeof(S::m);           // OK
int j = sizeof(S::m + 42);      // OK

— end example ]

未计算的操作数 where T::x可能出现可以是任何表达式。因此，即使是 C++20 之前的版本，您也可以编写

decltype(fun(T::x)) i{};