Why is st_blocks always reported in 512-byte blocks?

Question

I was debugging a fuse filesystem that was reporting wrong sizes for du. It turned out that it was putting st_size / st_blksize [*] into st_blocks of the stat structure. The Linux manual page for stat(2) says:

struct stat {
    …
    off_t     st_size;    /* total size, in bytes */
    blksize_t st_blksize; /* blocksize for file system I/O */
    blkcnt_t  st_blocks;  /* number of 512B blocks allocated */
    …
};

What is st_blksize for if st_blocks is in 512B blocks anyway?

[*] which looks wrong anyway, as integer division doesn't account for the fractional part…

By historical reason. All UNIXes use block (512 bytes) instead of kilobyte because in the past the size of disk sector was 512 bytes — Romeo Ninov, May 26 '19 at 15:25
You should put `st_blocks = (st_size + 511) / 512` -- maybe needlessly obfuscating it through an obtuse macro like `ROUNDING_UP_DIV_BY_SECTORS()` so the 1st midwit that comes along is impressed by the professional looking code instead of start carping about magic numbers and things he doesn't understand;-) — , May 27 '19 at 07:43
@RomeoNinov `In [4]: os.stat('/vmlinuz').st_blksize Out[4]: 4096` MAGIC! — LtWorf, Feb 22 '21 at 08:12

score 8 · Accepted Answer · answered May 27 '19 at 05:15

st_blocks is defined as

Number of blocks allocated for this object.

The size of a block is implementation-specific. On Linux it’s always 512 bytes, for historical reasons; in particular, it used to be the typical size of a disk sector.

st_blksize is unrelated; it’s

A file system-specific preferred I/O block size for this object. In some file system types, this may vary from file to file.

It indicates the preferred size for I/O, i.e. the amount of data that should be transferred in one operation for optimal results (ignoring other layers in the I/O stack).

Ah, so they have a different purpose actually. That explains a lot. Thanks! — liori, May 27 '19 at 08:49

Why is st_blocks always reported in 512-byte blocks?

1 Answers1