Doubts cast in c

In 64-bit architectures, pointers occupy 8 bytes (that is, sizeof(int*) == 8 ). On the other hand, integers have 4 bytes (that is, sizeof(int) == 4 ). This can be easily proven:

#include <stdio.h>

int main(void) {
    printf("%ld %ld", sizeof(int*), sizeof(int));
}

This code outputs 8 4 .

See here working on ideone.

When you take a pointer and do a cast for integer, you'll be throwing out 4 bytes of its value. This is not problematic per se, but picking up a memory address and throwing away a part of it is not an operation that makes much sense in practice and probably will give you incorrect values and no practical service, so the compiler gives you a < in> warning .

It is written in the notice itself. The type size is different.

Your int is probably 16-bit, and the long is 32-bit, and the pointer is 32-bit.

These sizes depend on the architecture.

Editing

It may be the case here, or not, of a little confusion, clarifying if this is the case: Declare

int *px = &x;

Creates a pointer that points to the address of the variable x, that is, px is a pointer. In the question you are converting a memory address to a int or long . To access the value stored in the address for which px points, you must use the "*", like this:

printf("%d %d\n", x, *px);

This will always print the same value twice. To print the value and then the address, use your very first example.

printf("%d %ld\n", x, (long)px);