[dllthomas]

"The n in strncpy describes to what size the destination buffer (not string) should be padded with '\0'."

This is false.

From the man page: "The strncpy() function is similar, except that at most n bytes of src are copied. Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated."

And the following code prints foo:ar on my system.

    char buffer[10];
    strcpy(buffer, "foobar");
    strcpy(buffer, "foo");
    printf("%s:%s\n", buffer, buffer+4);

Edited to add: huh, scratch that. Obvious error in above test :-P. Testing it with strncat like I had meant to, it seems it is in fact padded, not just (possibly) terminated. Interesting, and very worth knowing if you are trying to move a probably small string to a large buffer under time pressure.

[tobiasu]

What is false? I'm not talking about the source string not being terminated when it's too long. That's obvious and there are plenty posts about it in this thread.

Not sure what you want to say with the example code. Maybe swap it for strncpy and strlcpy and see whether that matches your expectations?

[dllthomas]

Yeah, my bad, had meant to use strncpy. My contention had been that it only zeroes the first following character (if that). On closer reading of the man page, it is in fact clear.