|
Why does strncpy not always place a \0 terminator in the destination string?
C Interview Questions and Answers
(Continued from previous question...)
Why does strncpy not always place a \0 terminator in the destination string?
strncpy was first designed to handle a now-obsolete data structure, the fixed-length, not-necessarily-\0-terminated ``string.'' strncpy is admittedly a bit cumbersome to use in other contexts, since you must often append a '\0' to the destination string by hand.
You can get around the problem by using strncat instead of strncpy. If the destination string starts out empty (that is, if you do *dest = '\0' first), strncat does what you probably wanted strncpy to do:
*dest = '\0';
strncat(dest, source, n);
This code copies up to n characters, and always appends a \0.
Another possibility is
sprintf(dest, "%.*s", n, source)
(though, strictly speaking, this is only guaranteed to work for n <= 509).
When arbitrary bytes (as opposed to strings) are being copied, memcpy is usually a more appropriate function to use than strncpy.
(Continued on next question...)
Other Interview Questions
| 