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 *’
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 reference to a pointer, always means the same thing. you are just overthinking it (a sign of intelligence) . look at it like this, int *myintptr; means that the address the pointer myintptr is referencing will store an int, and the *myintptr = 5; means that the address the pointer myintptr is referencing should contain a 5. the * always means "the value at".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);
*name is derefencing the pointer (ie returning the thing that the pointer points to)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]);
}
}
