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 want to display a single character then you can also use name[0] instead of using pointer.

It will serve your purpose but if you want to display full string using %c, you can try this:

#include<stdio.h>
void main()
{ 
    char name[]="siva";
    int i;
    for(i=0;i<4;i++)
    {
        printf("%c",*(name+i));
    }
} 

Answer 2

The thing is that the printf function needs a pointer as parameter. However a char is a variable that you have directly acces. A string is a pointer on the first char of the string, so you don't have to add the * because * is the identifier for the pointer of a variable.

Answer 3

The name of an array is the address of its first element, so name is a pointer to memory containing the string "siva".

Also you don't need a pointer to display a character; you are just electing to use it directly from the array in this case. You could do this instead:

char c = *name;
printf("%c\n", c);

Answer 4

You're confusing the dereference operator * with pointer type annotation *. Basically, in C * means different things in different places:

• In a type, * means a pointer. int is an integer type, int* is a pointer to integer type
• As a prefix operator, * means 'dereference'. name is a pointer, *name is the result of dereferencing it (i.e. getting the value that the pointer points to)
• Of course, as an infix operator, * means 'multiply'.

Answer 5

%c

is designed for a single character a char, so it print only one element.Passing the char array as a pointer you are passing the address of the first element of the array(that is a single char) and then will be printed :

s

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

will print

i

and so on ...

Pointer is not needed for the %s because it can work directly with String of characters.

Answer 6

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]').