C Programming - Strings - Discussion

#include <stdio.h>
#include <string.h>

main()
{
char str[]="%f%d%c\n\t%d";
int i;
for(i=0;i<10;i++)
printf("%c",str[i]);
}

Output :

%f%d%c
	%d

Why no output is coming for this code ?

#include<stdio.h>
int xstrlen (char *);
void xstrcat (char*,char*,int );
main ()
{
char source []="hello";
char target [20]="world";
int length= xstrlen(target);
xstrcat(target,source,length);
printf("source string=%s",source);
printf("target string=%s",target);
}

int xstrlen (char *t)
{
int len=0;
while (*t!='\0')
len++;
return (len);

}

void xstrcat (char *t,char *s,int l)

{

while (*s!='\0')
{
t=t+l;
*t=*s;
t++;
s++;
}
*t='\0';
}

But in strings strnset is not a predefined functions. By compiling the above program it showing some error.

Can you explain?

This program gives run time error while executing.

What is the difference between strnset and strcset?

@Tejaswi you are right,

Executing this program, compiler gives following error.

In function 'main': Undefined reference to 'strnset'

What means for the strnset? and where to use strnset?

It sets the portion of characters in a string to given character.
String before strnset: abcdefghijklmnopqrstuvwxyz

String after strnset: xxxxxxxxxxxxxnopqrstuvwxyz

The user want to replace abcdefghijklm by using 'x' and given position is 13 .so 'a' to 'm' is replaced by 'x'. finally got xxxxxxxxxxxxxnopqrstuvwxyz.
and one thing

Use this program:

#include <stdio.h>
#include <string.h>

int main(void)
{
char string[70] = "abcdefghijklmnopqrstuvwxyz";
char letter = 'x';
printf("string before strnset: %s\n", string);
strnset(string, letter, 13);
printf("string after strnset: %s\n", string);
return 0;
}

As far as I know. We can't modify a string literal. Then how its getting modified without giving any error?

Can anyone explain?

Note: GCC Compiler (32 Bit Linux Platform).

#include<stdio.h>
#include<string.h>
int main()
{
char s[90]="endl";
printf("%s",strcat (s,strcat(s,s)));
return 0;
}

Why it shows segment fault?

Please, anyone explain me.