“需要"忽略字段不是静态的 [英] "requires" ignores a field is not static

问题描述

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(如预期的那样，因为 x 在 T 中不是静态的).但是，当我仅注释A行时，代码在Clang 11.0.0和GCC 10.2.0(都输出 1 )中都可以正常编译.我缺少 requires 是什么?它不应该返回 false 吗?

If I only comment Line B, then the code cannot be compiled (as expected, since x is not static in T). But when I only comment Line A, the code compiles just fine with both Clang 11.0.0 and GCC 10.2.0 (both output 1). What it is that I am missing about requires? shouldn't it return false?

推荐答案

要求表达式是一大堆未评估的操作数.

A requires expression is one big list of unevaluated operands.

[expr.prim.req]

2 require-expression是类型为bool的prvalue，其值为如下面所描述的.出现在需求主体中的表达式是未评估的操作数.

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总是可能出现在未评估的操作数中.

And a qualified-id naming a non-static data member always could appear in an unevaluated operand.

[expr.prim.id]

2 表示以下内容的id表达式非静态数据成员或类的非静态成员函数只能使用:

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

• 作为类成员访问的一部分，其中对象表达式引用成员的类或从该类派生的类，或者

• 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

形成指向成员([expr.unary.op])的指针，或

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

如果该id表达式表示一个非静态数据成员，并且它出现在未评估的操作数中.[示例:

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

—结束示例]

可能出现 T :: x 的未经评估的操作数可以是任何表达式.因此，即使在C ++ 20之前的版本中，您也可以编写

The unevaluated operand where T::x may appear can be any expression. So even pre-C++20 you could for example write

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

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