Copying strings in C

Question

I'm learning about strings in c. I am using code::blocks as a compiler, even though it's not just for c. So, the problem with the code below is that the output for string2 is the stored 5 characters plus string1's output. I'll show you:

#include <stdio.h>
#include <string.h>          /* make strncpy() available */

int main()
{
char string1[]= "To be or not to be?";
char string2[6];

/* copy first 5 characters in string1 to string2 */
strncpy (string2, string1, 5);

printf("1st string: %s\n", string1);
printf("2nd string: %s\n", string2);
return 0;
}

Output is:

1st string contains: To be or not to be? 
2nd string contains: To be To be or not to be?

If you ask me, that's a lot more than 5 characters...

Answer 1

From the strncpy man page:

No null-character is implicitly appended to the end of destination, so destination will only be null-terminated if the length of the C string in source is less than num.

Since the original string is greater in length than 5, no NULL is being added.

As others have pointed out, to add some safety to it:

strncpy (string2, string1, sizeof(string2));
string2[sizeof(string2)-1] = '\0';

Note that if string2 is obtained through a malloc():

char * string2 = malloc(123); //sizeof(string2) == sizeof(void *) and not 123

And the above code would fail.

For the sake of completeness, here is the code: http://ideone.com/eP4vd

Answer 2

You are not terminating string2 with '\0', so the printf overruns. Try to do:

memset(string2,0,6);

before using string2. or, since you know you are copying 5 chars, after strncpy:

string2[5] ='\0';

so you properly terminate correctly the string.

pay attention to the fact that you should put '\0' after exactly the number of characters you did copy, otherwise you will see garbage even in the middle of the string.