Thread: Problems with strncat

Hi there,

I am a newbie to c programming, in fact I started yesterday. I am facing a problem which I can't understand.

Given the following code:

Code:

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

int main(void)
{
	char* buffer = (char *)malloc(sizeof(char)*6);
	
	strncat(buffer, "ab", 2);
	strncat(buffer, "de", 2);
	
	printf(buffer);
	
	free(buffer);
	
	return 1;
}

Output:
# gcc -Wall -ansi test.c -o test
# ./test
abde

However, if I change the two strncat lines to...

Code:

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

int main(void)
{
	char* buffer = (char *)malloc(sizeof(char)*6);
	
	strncat(buffer, "%s", 2);
	strncat(buffer, "%s", 2);
	
	printf(buffer);
	
	free(buffer);
	
	return 1;
}

# gcc -Wall -ansi test.c -o test
# ./test
Segmentation fault

Question: Why segmentation fault? What's so special about %s that causes this error?

Thanks in advance.

Regards,
CashCow01

Originally Posted by laserlight

Well, one problem with the second example is that you are missing two arguments to printf that should correspond to the "%s%s" format string.

One problem with both examples is that you do not append a null character after using strncat().

Well, I believe the two arguments are not compulsory. Suppose I want to print the literal string "%s%s" without any format. How do I do that?

Thanks for pointing out the null character though.

Code:

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

int main(void)
{
	char* buffer = (char *)malloc(sizeof(char)*6);
	
	strncat(buffer, "%s%s\0", 4);
	
	printf(buffer);
	
	free(buffer);
	
	return 1;
}

# gcc -Wall -ansi test.c -o test
# ./test
Segmentation fault

Originally Posted by laserlight

They are not compulsory, but leaving them out results in undefined behaviour. So, they are compulsory unless you want a potential bug.


With printf, you would do this:

Code:

printf("%%s%%s");

Thanks laserlight . Didn't know that % is an escape character by itself.

Originally Posted by CashCow01

Code:

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

int main(void)
{
	char* buffer = (char *)malloc(sizeof(char)*6);
	
	strncat(buffer, "%s%s\0", 4);
	
	printf(buffer);
	
	free(buffer);
	
	return 1;
}

# gcc -Wall -ansi test.c -o test
# ./test
Segmentation fault

Your code is still bad/flawed even after the proposed fixes.
First of all, after allocation buffer can contain anything. Specifically, the first character doesn't need to be 0, meaning that the first strncat won't actually work.
Also, don't cast malloc in C.
Finally, sizeof(char) is defined to be 1, so you don't need to multiply 6 by sizeof(char).