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?

The name of a character array, like any array name, yields the address of the first element of the array. Therefore, the following holds for the array m1: m1 == &m1[0] , *m1 == 'L', and *(m1+1) == m1[1] == 'i' from C Primer Plus- STephen Prata , confused me too. — sidchelseafan
– sidchelseafan, Commented Dec 18, 2013 at 13:10
What gave you the idea that a char is the same as the address of a char (char *)?! I've explained this in an answer to one of your other questions, didn't I? — Elias Van Ootegem
– Elias Van Ootegem, Commented Dec 18, 2013 at 13:13
You have misread the equivalences. Apparently, the array is char m1[] = "Like"; or something similar. Saying that m1 == &m1[0] is accurate. Saying that *m1 == 'L' is equivalent to saying m1[0] == 'L' and given the initialization I showed is accurate. Similarly, *(m1+1) == 'i' and m1[1] == 'i' are individually accurate, but the composite statement *(m1+1) == m1[1] == 'i' does not evaluate to true in C, though it makes sense speaking loosely. — Jonathan Leffler
– Jonathan Leffler, Commented Dec 18, 2013 at 13:16

MOHAMED · Accepted Answer · 2013-12-18 13:12:07Z

4

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"

edited Dec 18, 2013 at 13:12

answered Dec 18, 2013 at 13:06

MOHAMED

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Comments

Jonathan Leffler · Accepted Answer · 2013-12-18 13:20:15Z

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.

edited Dec 18, 2013 at 13:20

answered Dec 18, 2013 at 13:08

Jonathan Leffler

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2 Comments

Jonathan Leffler Over a year ago

Yes, %p would work; it would do a completely different job, of course. %p will print the address; %s will print the substring. Of the two, the substring is probably more interesting to the OP, but you could add a printf("%p\n", (void *)&heart[i]); to the list of informative printing operations.

Elias Van Ootegem Over a year ago

@MarounMaroun: %p would also require a (void *) cast

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