C ++：嵌套结构的平面初始化列表？

Question

Having defined

struct A {
  int a,b;
};

struct B {
  A a;
  int b;
};

the following initializations are obvious:

B b1 = { { 1 } };    // initializes b1.a.a
B b2 = { { 1, 2 } }; // initializes b1.a.a, b1.a.b
B b3 = { { 1 }, 2 }; // initializes b1.a.a, b1.b

But I am suprised that VC++ 2013 also allows these initializations without any warning:

B b4 = { 1 };       // initializes b4.a.a
B b5 = { 1, 2 };    // initializes b5.a.a, b5.a.b
B b6 = { 1, 2, 3 }; // initializes b6.a.a, b6.a.b, b6.b

Are flat initializer lists for nested structures/classes standard C++?

Answer 1

Yes, this is standard C++. In aggregate initialization (and only in aggregate initialization, not in other forms of list-initialization), braces can be elided, effectively flattening the containment hierarchy of an aggregate.

Answer 2

Yes, it's a standard C++ which is described in the 8.5 Initializers .

It's called a brace elision and can be applied in the case of aggregate initialization.

From the n4296 , 8.5.1.12 Aggregates :

Braces can be elided in an initializer-list as follows. If the initializer-list begins with a left brace, then the succeeding comma-separated list of initializer-clauses initializes the members of a subaggregate; it is erroneous for there to be more initializer-clauses than members. If, however, the initializer-list for a subaggregate does not begin with a left brace, then only enough initializer-clauses from the list are taken to initialize the members of the subaggregate; any remaining initializer-clauses are left to initialize the next member of the aggregate of which the current subaggregate is a member.

Here is another example from the 8.5.1.13 :

struct A {
    int i;
    operator int();
};

struct B {
    A a1, a2;
    int z;
};

A a;
B b = { 4, a, a };

In this example a1 is initialized with 4 , a2 is initialized with a and z is initialized with the result of operator int() for a .