c++ - 具有私有(private)成员的结构的构造函数

Question

考虑以下代码:

class A
{
private:
    struct B { private: int i; friend class A; };

public:
    static void foo1()
    {
        B b;
        b.i = 0;
    }

    static void foo2()
    {
        B b = {0};
    }
};

为什么 foo1 有效而 foo2 无效？类 A 的结构初始化构造函数不是可见的吗？有没有办法在 C++11 中完成这项工作？

(注意，删除 private 使 foo2 工作。)

Answer 1

Why does foo1 work but not foo2? Is not the struct initializer constructor visible for class A?

B b = {0};

不起作用，因为 B 不是聚合。而且它不是聚合，因为它有一个非静态私有(private)数据成员。如果删除私有(private)说明符，B 将成为聚合，因此可以以这种方式进行初始化。

---

C++03 标准 8.5.1 聚合
第 7 段:

If there are fewer initializers in the list than there are members in the aggregate, then each member not explicitly initialized shall be value-initialized (8.5). [Example:

 struct S { int a; char* b; int c; };
 S ss = { 1, "asdf" };

initializes ss.a with 1, ss.b with "asdf", and ss.c with the value of an expression of the form int(), that is,0. ]

C++03 标准 8.5.1 §1:

An aggregate is an array or a class (clause 9) with no user-declared constructors (12.1), no private or protected non-static data members (clause 11), no base classes (clause 10), and no virtual functions (10.3).