Getting the size of member variable

Question

If there is a POD structure, with some member variables, for example like this:

struct foo
{
   short a;
   int b;
   char c[50];
   // ...
};

Answer 1

Use the obvious:

sizeof( foo::a )

In C++, sizeof is ALWAYS evaluated at compile time, so there is no runtime cost whatsoever.

Answer 2

You can do that in C++0x:

sizeof(foo::a);

Answer 3

C++-0x allows you to do this:

  std::cout << sizeof( foo::a ) << std::endl;
  std::cout << sizeof( foo::b ) << std::endl;
  std::cout << sizeof( foo::c ) << std::endl;

C++-0x allows sizeof to work on members of classes without an explicit object.

The paper is here: Extending sizeof to apply to non-static data members without an object (revision 1)

I saw the post about this above too late. Sorry.

Answer 4

5.3.3/1:

The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is not evaluated, or a parenthesized type-id.

The above means the following construct is well defined:

sizeof( ((foo *) 0)->a);

Answer 5

You can create macro wrapper of what @Erik suggested as:

#define SIZE_OF_MEMBER(cls, member) sizeof( ((cls*)0)->member )

And then use it as:

cout << SIZE_OF_MEMBER(foo, c) << endl;

Output:

50

Demo : http://www.ideone.com/ZRiMe

Answer 6

struct foo
{
   short a;
   int b;
   char c[50];
   // ...
   static const size_t size_a = sizeof(a);
   static const size_t size_b = sizeof(b);
   static const size_t size_c = sizeof(c);
};

Usage:

foo::size_a