Strings and character with printf

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 *’ 

7 Answers

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.

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'.

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); 

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)); } } 

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.

Or you can use %c like in the below code:

#include <stdio.h> void main() { char name[]="siva"; //prints as string form printf("%s\n",name); //print each letter on different line int size= sizeof(name); int i; for(i=0;i<size;i++){ printf("%c\n",name[i]); } }