c++ - 将默认赋值运算符声明为 constexpr : which compiler is right?

Question

考虑

struct A1 {
    constexpr A1& operator=(const A1&) = default;
    ~A1() {}
};
struct A2 {
    constexpr A2& operator=(const A2&) = default;
    ~A2() = default;
};
struct A3 {
    ~A3() = default;
    constexpr A3& operator=(const A3&) = default;
};

GCC 和 MSVC 接受所有三个结构。 Clang 拒绝 A1 和 A2(但接受 A3)，并带有以下错误消息:

<source>:2:5: error: defaulted definition of copy assignment operator is not constexpr
    constexpr A1& operator=(const A1&) = default;
    ^
<source>:6:5: error: defaulted definition of copy assignment operator is not constexpr
    constexpr A2& operator=(const A2&) = default;
    ^
2 errors generated.

( live demo )

哪个编译器是正确的，为什么？

Answer 1

我认为这三个编译器都错了。

[dcl.fct.def.default]/3说:

An explicitly-defaulted function that is not defined as deleted may be declared constexpr or consteval only if it would have been implicitly declared as constexpr. If a function is explicitly defaulted on its first declaration, it is implicitly considered to be constexpr if the implicit declaration would be.

复制赋值运算符何时隐式声明constexpr？ [class.copy.assign]/10 :

The implicitly-defined copy/move assignment operator is constexpr if

• X is a literal type, and
• [...]

文字类型在哪里，来自 [basic.types]/10 :

A type is a literal type if it is:

• [...]

• a possibly cv-qualified class type that has all of the following properties:

  • it has a trivial destructor,
  • [...]

A1 没有普通的析构函数，因此它的隐式复制赋值运算符不是 constexpr。因此，复制赋值运算符格式不正确(gcc 和 msvc 错误接受)。

另外两个没问题，拒绝A2是个clang bug。

---

注意我引用的 [dcl.fct.def.default] 的最后一点。如果您明确默认，您实际上不必添加 constexpr 。在可能的情况下，这将是隐含的 constexpr 。