strncpy() does not guarantee that the copied string would be terminated with a null byte ('\0'). For a call that looks like strncpy(dst, src, n), if there is no null byte in the first n bytes of src, the string copied to dst would also not contain a null byte.

Here is an example code to demonstrate the problem:

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

  int main()
  {
      char a[] = "01234567";
      strncpy(a, "foobar", 4);
      printf("%.8s\n", a);
      return 0;
  }

Here is the output:

  $ cc -std=c89 -Wall -Wextra -pedantic foo.c && ./a.out
  foob4567

A C89-conforming alternative I use is a macro like this that guarantees '\0'-termination as the first thing:

  #define strcp(a, b, c) (a[0] = '\0', strncat(a, b, c - 1))

An example from my code: https://github.com/susam/uncap/blob/master/uncap.c#L78-L87

Here is how to use it:

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

  #define strcp(a, b, c) (a[0] = '\0', strncat(a, b, c - 1))

  int main()
  {
      char a[] = "01234567";
      strcp(a, "foobar", 4);
      printf("%.8s\n", a);
      return 0;
  }

Here is the output:

  $ cc -std=c89 -Wall -Wextra -pedantic foo.c && ./a.out
  foo