评估文件的权限位

Question

在stat.h中有一些constants/definitions像：

#define S_IRWXU 0000700         /* RWX mask for owner */
#define S_IRUSR 0000400         /* R for owner */
#define S_IWUSR 0000200         /* W for owner */
#define S_IXUSR 0000100         /* X for owner */

我知道我可以使用 stat 得到 'everything':

$ stat file.txt
  File: file.txt
  Size: 5307        Blocks: 16         IO Block: 4096   regular file
Device: 10302h/66306d   Inode: 13129530    Links: 1
Access: (0775/-rwxrwxr-x)  Uid: ( 1000/  ubuntu)   Gid: ( 1000/  ubuntu)
Access: 2020-09-25 21:47:31.013446496 +0000
Modify: 2020-09-25 22:34:21.527460575 +0000
Change: 2020-09-25 22:34:21.531460600 +0000
 Birth: -

在 C 中，我如何访问文件的结构来检查文件是否具有其他可执行权限，相当于：

// if the file in question has executable permissions for 'other'
if (file.txt & S_IXOTH) {
    // do something
}

Answer 1

stat 是一个 Linux 系统调用，而不仅仅是一个 shell 命令。

man 2 stat

这意味着你可以这样写代码：

#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>

int main() {

    struct stat buf;

    if (0 == stat("file.txt", &buf)) {
        if (buf.st_mode & S_IXOTH) {
            /* do what you wanted */
        }
    } else {
        /* handle error accessing the file here **/
    }

    return 0;
}

评估文件的权限位

Evaluating permission bit of a file

c

linux

stat