Printing a pointers value

Question

#include 

int main(void)
{
   int x = 99;
   int *pt1;

   pt1 = &x;

   printf(\"Value at p1: %d\\n\", *pt1);
   printf(\"Address of p1 (with %%         


        

Answer 1

• sizeof( int * ) need not be equal to sizeof( int )

• the library implementation is free to use a different, more appropriate presentation of the pointer value (e.g. hexadecimal notation).

• %p is simply "the right thing to do". It's what the standard says should be used for pointers, while using %d is misusing the integer specifier.

Imagine what happens when someone refactors your code, and somewhat over-enthusiastically exchanges all "int" with "long" and all "%d" with "%ld"... oops.