Problem in pointer size!!

[Ratan]

In my assigment,I have a question:

"Between a long pointer and a char pointer,which one consumes more momery.Explain."

But i have read that all pointer have same size as of 2 bytes.So what can be the answer of this question??

[Banfa]

All pointers are platform dependent, the is no constraint in C for pointers to different types to be the same size but similarly the is no constraint that they be different sizes.

If (as stated) on your system all pointers are 2 bytes in size then that is the answer to the question, "which one consumes more memory?". neither they both consume the same amount of memory.

It is a trick question, your teacher is testing to see if you have understood that pointers are a different type to the things they point at which I think you have.

[ssehgal2010]

hi banta
i believe pointer's size depends upon their type
like for following code

#include<stdio.h>

main(){

char *ptr;
long *p;
printf("%d\t%d\n",sizeof(*ptr),sizeof(*p));

exit (2);
}

the output is
1 4

[Ratan]

Hello ssehgal2010 ,
Your code helped me:

#include<stdio.h>

void main(){

char *ptr;
long *p;
printf("%d\t%d\n",sizeof(*ptr),sizeof(*p));
printf("%d\t%d\n",sizeof(char *),sizeof(long *));

}
Output comes out to be:
1 4
2 2

Can i know the reasons why this has happen?

[cgorbit]

What's the trouble?

int n;
char c;

int * pn = &n;
char * pc = &c;

n - integer value in memory that contain 32b - 4bytes
c - integer/symbol value in memory that contain 8b - 1bytes

To point to any value you must know address of its first byte.
When you addressing byte you shoudn't worry about which type of value placed there. Therefore you can use equal variable to hold address. (on my IA-32 its allocate 4bytes like an integer value)

That's why pn and pc occupy the same amount of memory.

[Banfa]

char *ptr;
long *p;
printf("%d\t%d\n",sizeof(*ptr),sizeof(*p));

This is not printing the size of the pointers but the size of the types they are pointing to.

printf("%d\t%d\n",sizeof(char *),sizeof(long *));

this is printing the size of the pointers

When you addressing byte you shoudn't worry about which type of value placed there.

This is not actually true because C does not guarantee that pointers to different types have the same size or bit pattern. However on many (most?) systems today pointers to different types do have the same size and bit pattern. Sometimes programmers rely on this which in theory makes their code non-portable, however in practice this is not often a problem.

However just because it is not a problem doesn't mean you can ignore it, you should know it in case you happen to work on a system where pointers to different types are of non-equal size or bit pattern.

[nikes90]

Hey.. U are printing sizeof (*ptr) which is the variable pointed by the pointer
try to print sizeof(ptr) then you will get correct answer.

main()
{
char *cptr = "Abc";
int *iptr = 10;
printf(" %d\t%d\n",sizeof(cptr),sizeof(iptr));
}

out put:

2 2