如何在Posix系统中获得总可用磁盘空间？

Question

我正在编写一个跨平台的应用程序,我需要总的可用磁盘空间.对于posix系统(Linux和Macos),我使用的是statvfs.我创建了这个C++方法:

long OSSpecificPosix::getFreeDiskSpace(const char* absoluteFilePath) {
   struct statvfs buf;

   if (!statvfs(absoluteFilePath, &buf)) {
      unsigned long blksize, blocks, freeblks, disk_size, used, free;
      blksize = buf.f_bsize;
      blocks = buf.f_blocks;
      freeblks = buf.f_bfree;

      disk_size = blocks*blksize;
      free = freeblks*blksize;
      used = disk_size - free;

      return free;
   }
   else {
      return -1;
   }
}

不幸的是,我得到了一些我无法理解的奇怪价值观.例如:f_blocks = 73242188 f_bsize = 1048576 f_bfree = 50393643 ...

这些值是以位,字节还是其他形式存在的？我在这里读的stackoverflow那些应该是字节,但后来我得到的总字节数是:f_bsize*f_bfree = 1048576*50393643但这意味着49212.542GB ......太多了......

我是否对代码或其他任何内容做错了？谢谢!

Answer 1

I don't know OSX well enough to predict this is definitely the answer, but f_blocks and f_bfree actually refer to "fundamental blocks", or "fragments" (which are of size buf.f_frsize bytes), not the "filesystem block size" (which is buf.f_bsize bytes):

http://www.opengroup.org/onlinepubs/009695399/basedefs/sys/statvfs.h.html

f_bsize is just a hint what the preferred size is for I/O operations, it's not necessarily anything to do with how the filesystem is divided.

Answer 2

我认为最后两个答案是正确且有用的。不过，我通过简单地将函数statvfs替换为函数statfs来解决。块大小如预期的那样为 4096，一切似乎都是正确的。谢谢！