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Why shouldn't it? What did you expect?

both the size of int and sizeof ptr is 4 byte in 32 bit machine.so why size of ptr is 8 byte in 64 bit machine

Paste Code that does so !!

#include<stdio.h>
#include<stdlib.h>

int main()

{

int * ptr;
int x;

printf ( "sizeof ptr =%d size of int =%d ",sizeof(ptr) ,sizeof(int));

return 0;

}

on compilation:

In function ‘main’:
11:1: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat]
11:1: warning: format ‘%d’ expects argument of type ‘int’, but argument 3 has type ‘long unsigned int’ [-Wformat]

./a.out
sizeof ptr =8 size of int =4

changed the code to printf ( "sizeof ptr =%ld size of int =%ld ",sizeof(ptr) ,sizeof(int));

Now not getting any warning but output is same
sizeof ptr =8 size of int =4

I think, you meant gcc compiler on an 64bit m/c.
You may check for the __SIZEOF_INT__ and __SIZEOF_POINTER__ predefined macros in gcc, they are independent of OS and m/c.

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