Strings and character with printf

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 is

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 use

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

Error i got is

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]').