Question

I was confused with usage of %c and %s in the following C program:

#include <stdio.h>
    
void main()
{
    char name[] = "siva";
    printf("%s\n", name);
    printf("%c\n", *name);
}

Output:

siva
s

Why we need to use pointer to display a character %c, and pointer is not needed for a string

I am getting error when I run

printf("%c\n", name);

I got this error:

str.c: In function ‘main’:
str.c:9:2: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’

Answer 1

If you try this:

#include<stdio.h>

void main()
{
 char name[]="siva";
 printf("name = %p\n", name);
 printf("&name[0] = %p\n", &name[0]);
 printf("name printed as %%s is %s\n",name);
 printf("*name = %c\n",*name);
 printf("name[0] = %c\n", name[0]);
}

Output is:

name = 0xbff5391b  
&name[0] = 0xbff5391b
name printed as %s is siva
*name = s
name[0] = s

So ‘name’ is actually a pointer to the array of characters in memory. If you try reading the first four bytes at 0xbff5391b, you will see ‘s’, ‘i’, ‘v’ and ‘a’

Location     Data
=========   ======

0xbff5391b    0x73  's'  ---> name[0]
0xbff5391c    0x69  'i'  ---> name[1]
0xbff5391d    0x76  'v'  ---> name[2]
0xbff5391e    0x61  'a'  ---> name[3]
0xbff5391f    0x00  '\0' ---> This is the NULL termination of the string

To print a character you need to pass the value of the character to printf. The value can be referenced as name[0] or *name (since for an array name = &name[0]).

To print a string you need to pass a pointer to the string to printf (in this case name or &name[0]).