c - "int j = (strlen) (str)"是什么意思？

Question

我从 bstrlib 中找到了以下代码，在 bstrlib.c 中(例如第 193 行):

bstring bfromcstr (const char * str) {
    bstring b;
    int i;
    size_t j;

    if (str == NULL) return NULL;
    j = (strlen) (str);
    i = snapUpSize ((int) (j + (2 - (j != 0))));
    if (i <= (int) j) return NULL;

    b = (bstring) bstr__alloc (sizeof (struct tagbstring));
    if (NULL == b) return NULL;
    b->slen = (int) j;
    if (NULL == (b->data = (unsigned char *) bstr__alloc (b->mlen = i))) {
        bstr__free (b);
        return NULL;
    }

    bstr__memcpy (b->data, str, j+1);
    return b;
}

行 (strlen) (str) 是什么意思？这种使用strlen的代码在bstrlib.c中很常见。

我在 OSX 机器上做了一些测试代码，但我无法理解也无法解释它的含义。

Answer 1

j = (strlen) (str);本质上与 j = strlen(str); 相同.

按照他们的方式编写代码将确保真实 strlen函数(来自 <string.h> )将被使用，即使它被 function-like macro 隐藏名为 strlen() .

A function-like macro is only expanded if its name appears with a pair of parentheses after it. If you write just the name, it is left alone. This can be useful when you have a function and a macro of the same name, and you wish to use the function sometimes.

因为在 (strlen) (str) , strlen后面没有一对括号，宏不会展开。

这个例子应该展示:

#include <stdio.h>
#include <string.h>

/**
 * Only check something like this in if you're
 * planning on changing jobs very soon!
 */
#define strlen(s)   666

int main(void)
{
    const char *str = "Stack Overflow";
    size_t j;

    /* This always uses the *real* strlen */
    j = (strlen) (str);
    fprintf(stderr, "(strlen) (str) = %zu\n", j);

    /* This might be using a function-like macro */
    j = strlen(str);
    fprintf(stderr, "strlen(str) = %zu\n", j);

    return 0;
}

输出:

$ ./a.out 
(strlen) (str) = 14
strlen(str) = 666

当然，下面的方法也行得通:

#include ...
...

#ifdef strlen
#undef strlen
#endif