Is strncpy really less secure than strscpy ?

[James Dutton]

Is strncpy really less secure than strscpy ?

If one uses strncpy and thus put a limit on the buffer size during the
copy, it is safe. There are no writes outside of the buffer.
If one uses strscpy and thus put a limit on the buffer size during the
copy, it is safe. There are no writes outside of the buffer.
But, one can fit more characters in strncpy than strscpy because
strscpy enforces the final \0 on the end.
One could argue that strncpy is better because it might save the space
of one char at the end of a string array.
There are cases where strncpy might be unsafe. For example copying
between arrays of different sizes, and that is a case where strscpy
might be safer, but strncpy can be made safe if one ensures that the
size used in strncpy is the smallest of the two different array sizes.

If one blindly replaces strncpy with strscpy across all uses, one
could unintentionally be truncating the results and introduce new
bugs.

The real insecurity surely comes when one tries to use the string.
For example:

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

int main() {
        char a[10] = "HelloThere";
        char b[10];
        char c[10] = "Overflow";
        strncpy(b, a, 10);
        /* This overflows and so in unsafe */
        printf("a is  %s\n", a);
        /* This overflows and so in unsafe */
        printf("b is  %s\n", b);
        /* This is safe */
        printf("b is  %.*s\n", 10, a);
        /* This is safe */
        printf("b is  %.*s\n", 4, a);
        return 0;
}


So, why isn't the printk format specifier "%.*s" used more instead of
"%s" in the kernel?

Kind Regards

James