Why should you use strncpy instead of strcpy?

Question

Edit: I've added the source for the example.

I came across this example:

char source[MAX] = "123456789";
char source1[MAX] = "123456789";
char destination[MAX] = "abcdefg";
char destination1[MAX] = "abcdefg";
char *return_string;
int index = 5;

/* This is how strcpy works */
printf("destination is originally = '%s'\n", destination);
return_string = strcpy(destination, source);
printf("after strcpy, dest becomes '%s'\n\n", destination);

/* This is how strncpy works */
printf( "destination1 is originally = '%s'\n", destination1 );
return_string = strncpy( destination1, source1, index );
printf( "After strncpy, destination1 becomes '%s'\n", destination1 );

Which produced this output:

destination is originally = 'abcdefg'
After strcpy, destination becomes '123456789'

destination1 is originally = 'abcdefg'
After strncpy, destination1 becomes '12345fg'

Which makes me wonder why anyone would want this effect. It looks like it would be confusing. This program makes me think you could basically copy over someone's name (eg. Tom Brokaw) with Tom Bro763.

What are the advantages of using strncpy() over strcpy()?

Answer 1

The strncpy() function was designed with a very particular problem in mind: manipulating strings stored in the manner of original UNIX directory entries. These used a fixed sized array, and a nul-terminator was only used if the filename was shorter than the array.

That's what's behind the two oddities of strncpy():

• It doesn't put a nul-terminator on the destination if it is completely filled; and
• It always completely fills the destination, with nuls if necessary.

For a "safer strcpy()", you are better off using strncat() like so:

if (dest_size > 0)
{
    dest[0] = '\0';
    strncat(dest, source, dest_size - 1);
}

That will always nul-terminate the result, and won't copy more than necessary.