具有私有成员的Struct的构造函数

Question

Consider the following code:

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};
    }
};

Why does work but not ? Is not the struct initializer constructor visible for class A? Is there anyway to make this work in C++11?

(Note, removing the private makes foo2 working.)

Answer 1

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

B b = {0};

Does not work because B is not an Aggregate . And it is not an Aggregate because it has an non-static private data member. If you remove the private specifier, B becomes an Aggregate and hence can be initialized in this manner.

---

C++03 Standard 8.5.1 Aggregates
Para 7:

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 standard 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).