Display filesystem's free space available to the root user

12

3

By default, Linux reserves some fraction of a filesystem for the root user, at least on ext4 (to prevent ordinary users from filling the drive completely and crashing the system, and to reduce fragmentation). However, df only displays the free space apparent to regular users (even when run as root). How do you display the "real" free space, that would be accessible to root?

Mechanical snail

Posted 2012-07-03T05:56:50.033

Reputation: 6 625

Answers

6

I am not sure there is a tool built in, but assuming you have left the reserved at the default 5% then this will tell you:

df / | grep dev | cut -f 3,6 -d\  | awk '{print ($1*.05)+$2}'

df the root, grep for the line with dev in it (to remove the header), cut the size and available fields, then use an awk script to calculate 5% of the disk size added to the available.

You could pull the actual reservation from tune2fs -l <device> and combine this with the above in a script.

Paul

Posted 2012-07-03T05:56:50.033

Reputation: 52 173

2Right, so the following command will tell us how much of the partition space is reserved for privilegied users: tune2fs -l /dev/DEVICE | egrep "Block count|Reserved block count". E.g. for my "multimedia buffer" partition : Block count: 2621440 and Reserved block count: 128449 : 4.9% of the available blocks (conservative setting to help prevent fragmentation). – tuk0z – 2016-07-09T13:18:57.820

There is no need to use the cut command: you can pick the fields by awk directly. In fact the cut command did not work for me as expected. Anyway, I gave another answer that should give more precise answer. – jarno – 2017-04-24T18:00:32.290

1Execute tune2fs -m <percentage> <device-name> to change disk space reserved for only for root usage. – luka5z – 2017-05-08T10:15:32.990

1Inspired by the above I came up with this to calculate reserved space on root volume in MB: sudo tune2fs -l $(df | grep -E '/$' | cut -d\ -f 1) | egrep "Reserved block count|Block size" | paste -sd\ | awk '{print ($4 * $7 / ( 1024 * 1024 ) ), "MB"}' – mp3foley – 2018-09-19T22:57:01.577

This is not a portable answer. Check below for the answer you need (spoiler: tune2fs /dev/sda1 or even better stat -f -c '%a blocks free and %f blocks free for root (%S bytes per block)' / – trs – 2019-06-24T01:03:08.547

17

By using the command tune2fs (found in /sbin/tune2fs), you can easily determine the reserved space: (and more!)

tune2fs -l /dev/sda1

I'll provide my system's info for reference, I'm going to remove extraneous lines not important to this question:

The header... and volume name, I label all my drives, makes them easy to identify if needed.

tune2fs 1.42.4 (12-Jun-2012)
Filesystem volume name:   xenon
Last mounted on:          /
...

REALLY want this to say "clean" while the system is running. Honest!

Filesystem state:         clean

This is where the data storage capacity information begins:

Here you can see that I have 121,179,648 blocks total... with a block size of 4K (4096), that multiplies out to some big number (462-ish GB). (Block size is noted below)

Block count:              121179648

And the reserved blocks... by looking at the number above, and the number below.. you should be able to relatively quickly figure out I have 1% reserved. In this case (4.62-ish GB)

Reserved block count:     1211796

How much free space currently available? Right here!

Free blocks:              104090586
...

And the all important block size. Useful for multiplying.

Block size:               4096
...

These lines say WHO the blocks are reserved for... user 0, root, in this case

Reserved blocks uid:      0 (user root)
Reserved blocks gid:      0 (group root)
...

There's lots more information available here, but this should give you an ability to quickly ascertain how much is available, and how much more is reserved for root. Simple math.

Hope this helps. Remember...man pages are your friends.

lornix

Posted 2012-07-03T05:56:50.033

Reputation: 9 633

2

This displays free space in bytes in partition related to "/path"

printf '%s' $(($(stat -f --format='%f*%S' /path)))

You do not have to be superuser to run it.

BTW I do not quite understand what is the difference between

%s block size (for faster transfers)

%S fundamental block size (for block counts)

in usage of stat.

jarno

Posted 2012-07-03T05:56:50.033

Reputation: 173

Cool - does this account for the space reserved for root? – Paul – 2017-04-25T13:29:12.070

@Paul yes, it is included – jarno – 2017-04-25T21:11:59.237

Your output is effectively a sum - so you could pipe the output into bc to get the byte count, or even use --format='%f*%S/1024/1024/1024' | bc to get kilobytes / megabytes / gigabytes depending on how many divisions you choose. – Paul – 2017-04-26T01:18:15.780

@Paul Well, I'd call it a product. The binary prefixes are called kibibytes, mebibytes and gibibytes, respectively. The surrounding arithmetic expansion $(( ... )) does the calculation, but it results only the integer part of the quotient (if you add division). Or do you mean your shell does not support arithmetic expansion?

– jarno – 2017-04-26T07:41:05.763

@Paul But, if you use bc, instead, you could get the quotient with arbitrary precision, if you need that. – jarno – 2017-04-26T07:55:42.907

Ooops, crap you are right, I missed the expansion! – Paul – 2017-04-26T23:27:12.847

This is great. Thanks. Have been looking for this for a very long time. THIS should be the accepted answer as it is portable and easy to use. – trs – 2019-06-24T01:03:31.653