How do I programmatically get the free disk space for a directory in Linux
Solution 1:
check man statvfs(2)
I believe you can calculate 'free space' as 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 */
};
Solution 2:
You can use 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);
Example:
#include <boost/filesystem.hpp>
using namespace boost::filesystem;
space_info si = space(".");
cout << si.available << endl;
Returns: An object of type space_info. The value of the space_info object is determined as if by using POSIX statvfs() to obtain a POSIX struct statvfs, and then multiplying its f_blocks, f_bfree, and f_bavail members by its f_frsize member, and assigning the results to the capacity, free, and available members respectively. Any members for which the value cannot be determined shall be set to -1.