Why char size increases as instance variable

Question

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Why isn’t sizeof for a struct equal to the sum of sizeof of each member?
How is the size of a C++ class determined?

When I check the size of the class with single char variable, it's size is 1 Byte. But if we add an integer, suddenly it goes to 8. Can you please explain Why?

class Char
{
  char b;
};
class Int
{
  int a;
};
class A
{
  int a;
  char b;
};

int main()
{
  Char Cobj;
  cout<<"Char size: "<<sizeof(Cobj)<<endl;
  Int Iobj;
  cout<<"Int size: "<<sizeof(Iobj)<<endl;
  A aobj;
  cout<<"A size: "<<sizeof(aobj)<<endl;

  return 0;
}

The output is: Char size: 1 Int size: 4 A size: 8

http://stackoverflow.com/questions/14510711/how-is-the-size-of-a-c-class-determined — Sergey Vyacheslavovich Brunov, Jan 29 '13 at 09:37

score 1 · Accepted Answer · answered Jan 29 '13 at 09:35

Because of padding - 3 dummy bytes will be added after A::b.

This is done to properly align A in, say, an array - the first member of A is an int, which has to have a specific alignment (4 or 8 bytes probably). So if you have

A arrayOfA[10];

the objects themselves have to be aligned to 4 or 8 bytes.

Why char size increases as instance variable

1 Answers1