c++ - 为什么静态 lambda 成员的封闭类型不完整？

Question

我今天尝试做类似的事情。我很惊讶它没有编译。

struct Test {
//  v----- Remove me to compile
    //  /*
    static constexpr auto get_test1 = [](Test const& self) {
        return self.test; // error, Test is incomplete
    };
    // */

    // Handwritten version of the lambda
    struct {
        constexpr auto operator() (Test const& self) const {
            return self.test; // ok
        }
    }
    static constexpr get_test2{};

    int test;
};

Live example

它表示 Test 类型在范围内不完整。然而 lambda 的手写版本确实有效。这样做的技术原因是什么？是标准中的疏忽，还是有特定的措辞使 Test 在 lambda 中不完整？

Answer 1

这是我能找到的:

§5.1.2 Lambda expressions [expr.prim.lambda]

5. [...] [ Note: Names referenced in the lambda-declarator are looked up in the context in which the lambda-expression appears. —end note ]

6. The lambda-expression’s compound-statement yields the function-body (8.4) of the function call operator, but for purposes of name lookup (3.4), [...] the compound-statement is considered in the context of the lambda-expression.

如果我没有误读标准，这意味着参数以及正文中的 Test 和 self 都会在 lambda 的上下文中查看/考虑，这是 Test 不完整的类范围。

至于为什么允许嵌套类:

§9.2 Class members [class.mem]

2. A class is considered a completely-defined object type (3.9) (or complete type) at the closing } of the classspecifier . Within the class member-specification, the class is regarded as complete within function bodies, default arguments, using-declarations introducing inheriting constructors (12.9), exception-specification s, and brace-or-equal-initializer s for non-static data members (including such things in nested classes). Otherwise it is regarded as incomplete within its own class member-specification.