What is assignment via curly braces called? and can it be controlled?

Question

That is not assignment. That is initialization.

Such initialization is allowed for aggregate only, that includes POD class. POD means Plain Old Data type.

Example,

//this struct is an aggregate (POD class)
struct point3D
{
   int x;
   int y;
   int z;
};

//since point3D is an aggregate, so we can initialize it as
point3D p = {1,2,3};

See the above compiles fine : http://ideone.com/IXcSA

But again consider this:

//this struct is NOT an aggregate (non-POD class)
struct vector3D
{
   int x;
   int y;
   int z;
   vector3D(int, int, int){} //user-defined constructor!
};

//since vector3D is NOT an aggregate, so we CANNOT initialize it as
vector3D p = {1,2,3}; //error

The above does NOT compile. It gives this error:

prog.cpp:15: error: braces around initializer for non-aggregate type ‘vector3D’

See yourself : http://ideone.com/zO58c

What is the difference between point3D and vector3D? Just the vector3D has user-defined constructor, and that makes it non-POD. Hence it cannot be initialized using curly braces!

---

What is an aggregate?

The Standard says in section §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).

And then it says in §8.5.1/2 that,

When an aggregate is initialized the initializer can contain an initializer-clause consisting of a brace-enclosed, comma-separated list of initializer-clauses for the members of the aggregate, written in increasing subscript or member order. If the aggregate contains subaggregates, this rule applies recursively to the members of the subaggregate.

[Example:

struct A 
{
   int x;
   struct B 
   {
      int i;
      int j;
   } b;
} a = { 1, { 2, 3 } };

initializes a.x with 1, a.b.i with 2, a.b.j with 3. ]