Strings and character with printf

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'.

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]).


%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.

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