c++ - 为什么内联声明不是不完整的类型？

Question

考虑下面的代码:

struct Foo {
    struct Bar;
    Foo()
    {
        Bar bar; // Why isn't Bar an incomplete type?!
    }
    struct Bar {}; // Full definition
};

// struct Bar {}; // fails to compile due to incomplete type

int main()
{
    Foo foo;
}

它在至少 2 个编译器(gcc5.2、clang3.5)下编译得很好。我的问题是:

• 为什么 Bar 在构造函数 Foo::Foo 中不被视为不完整类型，因为我在构造函数上方将其前向声明，但在构造函数内部完全使用它?

每当我将 Foo::Bar 移出类，换句话说 Bar 成为一个独立的类时，我都会得到预期的

error: aggregate 'Foo::Bar bar' has incomplete type and cannot be defined

Answer 1

在成员规范中，类在函数体中被认为是完整的，来自草案 C++ 标准部分 9.2 [class.mem]:

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

这意味着您甚至不必转发声明 Bar ( see it live ):

struct Foo {
    Foo()
    {
        Bar bar; 
    }
    struct Bar {};  
};

前向声明有助于避免违反 3.3.7 paragraph 2 and 3 部分.