关于path alloc函数（APUE）

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于 2019-02-18 20:39:46 发布

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本文链接：https://blog.youkuaiyun.com/yffhhffv/article/details/87644487

本文深入探讨了C语言中用于路径分配的函数path_alloc的实现细节。该函数根据POSIX版本和系统配置动态确定路径的最大长度，并据此分配内存。文章通过代码示例展示了如何处理PATH_MAX未定义的情况，以及如何在不同POSIX版本下正确地分配路径字符串。

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                #include "apue.h"
#include <errno.h>
#include <limits.h>
#ifdef   PATH_MAX
static int   pathmax = PATH_MAX;
#else
static int   pathmax = 0;
#endif
#define SUSV3 200112L
static long posix_version = 0;
/* If PATH_MAX is indeterminate, no guarantee this is adequate */
#define PATH_MAX_GUESS   1024
char *
path_alloc(int *sizep) /* also return allocated size, if nonnull */
{
char *ptr;
int size;
if (posix_version == 0)
       posix_version = sysconf(_SC_VERSION);
if (pathmax == 0) {     /* first time through */
       errno = 0;
       if ((pathmax = pathconf("/", _PC_PATH_MAX)) < 0) {
         if (errno == 0)
            pathmax = PATH_MAX_GUESS; /* it's indeterminate */
         else
            err_sys("pathconf error for _PC_PATH_MAX");
       } else {
         pathmax++;    /* add one since it's relative to root */
       }
}
if (posix_version < SUSV3)
       size = pathmax + 1;
else
       size = pathmax;
if ((ptr = malloc(size)) == NULL)
       err_sys("malloc error for pathname");
if (sizep != NULL)
       *sizep = size;
return(ptr);
}

或
char*path_alloc(int* size)
{
char *p = NULL;
if(!size) return NULL;
p = malloc(256);
if(p)
*size = 256;
else
*size = 0;
return p;
}