I am sorting big files (>100Go), and to reduce time spent on disk writes, I am trying to use GNU sort's --compress-program parameter. (Related: How to sort big files?)

However, it appears in certain cases that only the first temporary is compressed. I would like to know why, and what I could do to compress all temporaries.

I am using:

  • sort (GNU coreutils) 8.25
  • lzop 1.03 / LZO library 2.09

Steps to reproduce the issue:

You will need something like ~15Go free space, ~10Go ram, some time

First, create a 10Go file with the following C code:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    unsigned long n;
    unsigned char i;
    for(n = 0; n < 1000000000; n++) {
        for(i = 0; i < 3; i++) {
            printf("%03d", rand() % 1000);
    return 0;

And running it:

$ gcc -Wall -O3 -o generate generate.c
$ ./generate > test.input  # takes a few minutes
$ head -n 4 test.input

Then, start the sort process:

$ LC_ALL=C sort -T . -S 9G --compress-program=lzop test.input -o test.output

After some time, suspend the process, and list the tempararies created in the same folder (due to -T .):

$ ls -s sort*
890308 sortc1JZsR
890308 sorte7O878
378136 sortrK37RZ
$ file sort*
sortc1JZsR: ASCII text
sorte7O878: ASCII text
sortrK37RZ: lzop compressed data - version 1.030, LZO1X-1, os: Unix

It seems that only sortrK37RZ (the first temporary created) has been compressed.

[Edit] Running that same sort command with -S set to 7G is fine (i.e. all temporaries are compressed) while with 8G the issue is present.

[Edit] lzop is not called for the other temporary

I tryied and used the following script as a wrapper for lzop:

set -e
echo "$$: start at $(date)" >> log
lzop $@
echo "$$: end at $(date)" >> log

Here is the content of the log file, when several temporaries are written to disk:

11109: start at Sun Apr 10 22:56:51 CEST 2016
11109: end at Sun Apr 10 22:57:17 CEST 2016

So my guess is that the compress program is not called at all.

  • perhaps only the first file has an lzop header and the others are continuation files (eg as from split), just to avoid huge files, so file cannot detect the type correctly.
    – meuh
    Apr 10, 2016 at 17:47
  • @meuh I looked at the content of the files, and the output from the file command is right. Which also makes sense when you look at the size of the temporaries (the ASCII ones being way bigger)
    – r0g
    Apr 10, 2016 at 18:28

1 Answer 1


Not reproduced here?

$ shuf -i1-10000000 > t.in
$ sort -S50M -T. t.in --compress-program=lzop  # ^z
$ file sort* | tee >(wc -l) > >(grep -v lzop)
$ fg   # ^c
$ sort --version | head -n1
sort (GNU coreutils) 8.25

What I'm guessing the issue is, is due to failure to fork() the compression process due to the large mem size, and then falling back to a standard write. I.E. sort(1) is using fork()/exec() when ideally it should be using posix_spawn() to more efficiently fork the compression process. Now fork() is CoW, but there is still overhead in preparation of the associated accounting structures for such a large process. In future versions of sort(1) we'll using posix_spawn() to avoid this overhead (glibc has only just got a usable implementation of posix_spawn() as of version 2.23).

In the meantime I would suggest to use a much smaller -S. Perhaps -S1G and below.

  • Thank you for your inputs. Your commands performed the same on my computer. You may well be right, with my example, I am not experimenting the issue with -S set to 7G, but the issue is present with 8G. It seems to be as if lzop is not called (see my edit). My computer is running 16G of ram if that's relevent. Using lower values for -S seems to be counter intuitive: my aim is to reduce the time for sorting, with a lot spent in disk IO… for which using more ram sounds like performance improvment!
    – r0g
    Apr 10, 2016 at 21:01
  • 1
    You were right! I downloaded the sort(1) source code and make it print the errno after fork(), and it shows errno=12, meaning that it cannot allocate memory. I guess I will have to wait for new version of sort(1) while using smaller -S in the meanwhile, and read fork(2) to understand why there is an issue. Many thanks!
    – r0g
    Apr 10, 2016 at 21:35
  • How safe you it be to set /proc/sys/vm/overcommit_memory to 1, since sort(1) is calling exec() just after forking? I tried it and it seems to do the trick, but I am not sure how stupid it would be to use that option.
    – r0g
    Apr 10, 2016 at 21:54
  • Yes overcommit_memory would work, but it's a system wide setting so can impact other things. As for smaller buffers incurring more disk I/O, it's worth testing but it might not impact too much. Another option to test is --batch-size (details in the info page). Another option to try is to specify -T to a RAM disk like /dev/shm etc. Apr 10, 2016 at 22:53
  • Thank you for your answer once again. So many things to test! It's always amazing when programs make sensible choices for your everyday use, but still allow to finely tune and change all the tradeoffs if you are an advanced user!
    – r0g
    Apr 11, 2016 at 16:33

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