如何在Linux中以编程方式获取目录的可用磁盘空间

Question

是否有一个函数可以返回给定目录路径的驱动器分区上的可用空间？

Answer 1

校验 man statvfs(2)

我相信您可以将“可用空间”计算为 f_bsize * f_bfree。

NAME
       statvfs, fstatvfs - get file system statistics

SYNOPSIS
       #include <sys/statvfs.h>

       int statvfs(const char *path, struct statvfs *buf);
       int fstatvfs(int fd, struct statvfs *buf);

DESCRIPTION
       The function statvfs() returns information about a mounted file system.
       path is the pathname of any file within the mounted file  system.   buf
       is a pointer to a statvfs structure defined approximately as follows:

           struct statvfs {
               unsigned long  f_bsize;    /* file system block size */
               unsigned long  f_frsize;   /* fragment size */
               fsblkcnt_t     f_blocks;   /* size of fs in f_frsize units */
               fsblkcnt_t     f_bfree;    /* # free blocks */
               fsblkcnt_t     f_bavail;   /* # free blocks for unprivileged users */
               fsfilcnt_t     f_files;    /* # inodes */
               fsfilcnt_t     f_ffree;    /* # free inodes */
               fsfilcnt_t     f_favail;   /* # free inodes for unprivileged users */
               unsigned long  f_fsid;     /* file system ID */
               unsigned long  f_flag;     /* mount flags */
               unsigned long  f_namemax;  /* maximum filename length */
           };