[SATYAN JANA]

/******************************/
class A
{
private:
  int a;
  char b;
};

class B
{
private:
  char b;
};

int main()
{
  A a;
  B b;
  cout<<sizeof(a)<<"  "<<sizeof(b)<<endl;
  return 0;
}
/******************************************/

What will be the output of the above program?
The output will be,-
8 1
Why?...
















In class B, 1 byte is as usual allocated for char 'b'. But in case of class A,
4 bytes for integer 'a' and next 4 bytes for char 'b'. This is because of packaging format.

[shabbir]

I didn't knew this. Its really kool so out of excitement I tried even the double one and it gave 16 and 1

[SATYAN JANA]

Hmmm,
even the size of pointers can change. If you have double and char* still it will be (8 + 8 =)16.

[shabbir]

Quote:

Originally Posted by SATYAN JANA

Hmmm,
even the size of pointers can change. If you have double and char* still it will be (8 + 8 =)16.

Yup, seen that.

[rahul_scss]

Can somebody please tell me what exactly the rules of this packaging format are???

[shabbir]

Quote:

Originally Posted by rahul_scss

Can somebody please tell me what exactly the rules of this packaging format are???

If you visualize the memory as rows and columns then if you need to allocate some memory in different rows[for different variables] then you have to allocate n no of cols of bytes for each variable and n is choosen as maximum size.

e.g.
Variable 1 mem [8 bytes] | | | | | | | |
Variable 2 mem [4 bytes] | | | |

Instead of 4 bytes it allocates 8 to keep the packaging format in good shape for second variable memory.

If you go about by sequential memory segment then also allocating each variable memory of n [choosen as maximum] is easier and so it allocates 8 + 8 bytes instead of 8 + 4 bytes.

[rahul_scss]

Thanks Very Much Shabbir...

[shabbir]

Quote:

Originally Posted by rahul_scss

Thanks Very Much Shabbir...

My pleasure.

[pradeep]

Very useful piece of information Shabbir bhai.