3 ms·

I usually use strncpy and enforce null at end, that virtually is the same that does a strlcpy. But if you know how long is the source and destiny buffers (and

• adekok 12y ago
  strncpy() zero-fills the array. Which seems overly wasteful. char buffer[1 << 24}] strncpy(buffer, "a", sizeof(buffer)); will set all of the array to zero, instead of just copying two bytes. And as you noted, it won't NUL terminate the string. Since strncpy() doesn't return C strings, it's not really a "C string" function. It's a horrible, inefficient bastardization which should be used by no one.
  • ploxiln 12y ago
    Meanwhile, strlcpy() returns the length of the source. So if you just want to copy a part of the source, or you're not sure if the source is terminated past the length you want to copy, it's either wasteful or dangerous.
    • wahern 12y ago
      Those are completely made-up objections. strlcpy also fails to fix me coffee in the morning, ergo it should be avoided.
  • Zardoz84 12y ago
    size_t len = (src_size < dst_size) ? src_size : dst_size; memcpy (dst, src, len); // or memmove dst[len-1] = 0; So not fills unecesary 0's and it's much more faster that any str?cpy
• noselasd 12y ago
  You rarely want to use strncpy. Keep in mind: 1) char buf[1024]; strncpy(buf, "hello", sizeof buf); now fills in 5 characters, and 1019 0 bytes in the buffer. (i.e. you're writing 1018 nul bytes that's usually not needed) 2) char buf[5]; strncpy(buf, "hello", sizeof buf); results in buf NOT being a string. (it isn't nul terminated). So with most use of strncpy, you'll have to also do buf[sizeof buf - 1] = 0;