difference between array and pointer notation of strings in C

Question

#include<stdio.h>
int main(void)
{

 char heart[]="I Love Tillie"; /* using array notation */

 int i;
 for (i=0;i<6;i++)
 {
   printf("%c",&heart[i]);  /* %c expects the address of the character we want to print     */
 }

 return 0;

}

If heart[i] and &heart[i] mean the same thing , which is the address of heart[i] , why is my program giving me this- ?????? , as output? Could someone please help me out here?

Answer 1

First of all

should be

printf("%c",heart[i]); // if you want to print the charachter

or

printf("%p",&heart[i]); // if you want to print the charachter address in the memory

and not

printf("%c",&heart[i])

The heart is an array of charachters and the heart[i] is the charachter number i in the array

The &heart[i] is the memory address of the element number i in the heart array. and to print the memory address you have to use "%p"

Answer 2

You are trying to print an address as a single character; this is bad news.

heart[i] is a single character; &heart[i] is the address of that character. They are not the same thing at all.

Try a loop like this:

for (i = 0; i < 6; i++)
{
     printf("%c", heart[i]);
     printf(": %s\n", &heart[i]);
}

See what a difference the different conversion specifications (and parameter types) make. If you wish, you can add a printf("%p ", (void *)&heart[i]); to the start of the loop to see how the address values change as you go through the loop.