c++ - 三路运算符 <=> 返回带有隐式转换函数的结构体

Question

考虑以下无用的代码:

struct S{
  constexpr operator int() const { return 0; }
  constexpr auto operator<=>(S) const { return *this; }
};

static_assert(S{} <= S{});

Clang 和 MSVC 接受此代码，但 GCC rejects它带有错误消息:

error: no match for 'operator<=' (operand types are 'S' and 'int')

哪个编译器是对的？如何 operator<=合成自 operator<=> ?

Answer 1

来自 [over.match.oper]( 3.4.1 和 8 ):

For the relational ([expr.rel]) operators, the rewritten candidates include all non-rewritten candidates for the expression x <=> y.

和

If a rewritten operator<=> candidate is selected by overload resolution for an operator @, x @ y is interpreted as [...] (x <=> y) @ 0 [...], using the selected rewritten operator<=> candidate. Rewritten candidates for the operator @ are not considered in the context of the resulting expression.

所以对于表达式 S{} <= S{}选定的运算符将为 S::operator<=>(S) const表达式将改写为 (S{} <=> S{}) <= 0 .在重写的表达式中，操作数的类型是 S和 int ，其中内置 operator<=(int, int)将被选中。所以最终表达式(在将 S 转换为 int 之后)将导致 0 <= 0 , 即 true .
总之，在这种情况下，Clang 和 MSVC 是正确的，而 GCC 似乎无法解释 (S{} <=> S{}) <= 0作为对内置操作符的调用(注意读取 operand types are 'S' and 'int' 的错误消息)。如果您更改 static_assert 中的条件成为重写的表达式 (S{} <=> S{}) <= 0 ，然后 all three compilers accept it .