Question

#include<stdio.h>

int main()
{
    int i = 10;
    int *p = &i;

    printf("\n address of initialized pointer p: %u \n", p);
    p = &(*p);
    printf("\n modified address of initialized pointer p:%u value:%d valuez address: %d \n", p, *p, &(*p));

    return 0;
}

the code outputs:-

address of initialized pointer p: 3221221820

modified address of initialized pointer p:3221221820 value:10 valuez address: -1073745476

Why is “&(*p)”, behaving differently when used in a assignment statement and in a printf statement ?

Update
Sorry, just format specifier mistake in printf ;).Thanks for the replies and pointing out.

Answer 1

You are using incorrect format specifier in printf. Using %d for printing addresses won’t work. Use %p rather. [%u for printing address isn’t correct either.]

This works as per expectation.