Splint:来自strcpy()的新存储？

Question

我正在尝试学习和更好地理解splint，我想知道我从这段代码中得到的一个错误：

#include <stddef.h>
#include <stdlib.h>
#include <string.h>

/*@null@*/ /*@only@*/ char *dupStr(const char *str) {
    char *copy;
    size_t len;

    len = strlen(str) + 1U;
    if (!(copy = malloc(len * sizeof *str))) {
        return NULL;
    }
    (void) strncpy(copy, str, len);
    return copy;
}

错误是：

Splint 3.1.2 --- 26 Feb 2013

test.c: (in function dupStr)
test.c:13:9: New fresh storage copy (type void) cast to void (not released):
                (void)strncpy(copy, str, len)
  A memory leak has been detected. Storage allocated locally is not released
  before the last reference to it is lost. (Use -mustfreefresh to inhibit
  warning)

Finished checking --- 1 code warning

正确的解决方案是将返回值分配给copy，而不是丢弃它(它消除了警告)吗？

Answer 1

你不想忽略strncpy的返回值，这就是splint抱怨的原因。你想要这样的东西：

if (strncpy(copy, str, len) == NULL)
    return NULL;