使用C ++ 17

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有多少空间？

Answer 1

检查man statvfs(2)

我相信你可以将{自由空间'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 */
           };

Answer 2

您可以使用boost :: filesystem：

struct space_info  // returned by space function
{
    uintmax_t capacity;
    uintmax_t free; 
    uintmax_t available; // free space available to a non-privileged process
};

space_info   space(const path& p);
space_info   space(const path& p, system::error_code& ec);

示例：

#include <boost/filesystem.hpp>
using namespace boost::filesystem;
space_info si = space(".");
cout << si.available << endl;

返回：类型为space_info的对象。 space_info对象的值确定为使用POSIX statvfs（）获取POSIX结构statvfs，然后将其f_blocks，f_bfree和f_bavail成员乘以其f_frsize成员，并将结果分配给容量，free和可用会员分别。任何无法确定其值的成员都应设置为-1。

Answer 3

使用C ++ 17

您可以使用std::filesystem::space：

#include <iostream>  // only needed for screen output

#include <filesystem>
namespace fs = std::filesystem;

int main()
{
    fs::space_info tmp = fs::space("/tmp");

    std::cout << "Free space: " << tmp.free << '\n'
              << "Available space: " << tmp.available << '\n';
}

Answer 4

可以使用如下管道将命令输出到程序中：

char cmd[]="df -h /path/to/directory" ;
FILE* apipe = popen(cmd, "r");
// if the popen succeeds read the commands output into the program with 
while (  fgets( line, 132 , apipe) )
{  // handle the readed lines
} 
pclose(apipe);
// -----------------------------------

如何以编程方式获取Linux中目录的可用磁盘空间

4 个答案:

使用C ++ 17