c strncpy null 是否终止

Question

这个问题在这里已经有了答案:





strncpy documentation question

(6 个回答)


4年前关闭。




我正在阅读 this文件，它说:

char *strncpy(char *destination, const char *source, size_t num);

Copy characters from string Copies the first num characters of source to destination. If the end of the source C string (which is signaled by a null-character) is found before num characters have been copied, destination is padded with zeros until a total of num characters have been written to it.

No null-character is implicitly appended at the end of destination if source is longer than num. Thus, in this case, destination shall not be considered a null terminated C string (reading it as such would overflow).

destination and source shall not overlap (see memmove for a safer alternative when overlapping).

但我对这句话感到困惑:

in this case, destination shall not be considered a null terminated C string (reading it as such would overflow)

自如果 num > strlen(source) ，它将填充 '\0'最后， '\0'实际上是字符串中的空(终止)字符，为什么不应将其视为空终止的 C 字符串？

我写了下面的代码来验证:

  char from[] = { 'h', 'e', 'l', 'l', 'o', '\0' };
  char to[1024];
  for (int i = 0; i < 1024; i++) {
      to[i] = 'e';
  }
  strncpy(to, from, 1024);
  printf("from %s\n", from);

它适用于以下输出:

from hello
to hello

Answer 1

说的是strlen(source)的情况> num .它只会复制 num字符，其中没有一个是 NUL 并且不会添加 NUL。