How to set nproc for a single command?

Question

If you have a test.sh containing

make -j$(nproc)

How could you temporarily adjust nproc, along the lines of nproc=1 ./test.sh?

Answer 1

Create a wrapper that prints $nproc:

#!/bin/sh
[ -z "$nproc" ] && exec /usr/bin/nproc "$@"
printf '%s\n' "$nproc"

Name it nproc, make it executable (chmod +x nproc) and place it in your /special/directory for custom wrappers. Now you have two options:

1. On demand:

   PATH="/special/directory:$PATH" nproc=1 ./test.sh
   

2. Permanent. Modify your PATH so /special/directory is always there as the first entry. Then you can just

   nproc=1 ./test.sh
   

Notes:

• If the script alters (resets, sanitizes) its PATH, the solution may not work.
• If the variable is empty or not set, the wrapper uses the real nproc (by its full path to avoid calling itself recursively). This way nproc will behave exactly as before, unless you set the variable to any non-empty value.

• In your example $(nproc) is not quoted. This means you can inject additional options via the variable (e.g. nproc="2 -i", the wrapper will print this string, in the script it will be split into two words and -i will be passed to make). You may

  • take advantage of this and use the variable to actually inject options (hacky, not recommended);
  • always make sure the variable is sane;

  • implement some logic into the wrapper, so its output is less likely to be split; example:

    set -- $nproc
    printf '%s\n' "$1"
    # word splitting done in the wrapper, only the first word is printed
    

  • (proper solution) fix the script:

    make -j"$(nproc)"
    # if the wrapper prints unexpected string, make will comply
    

Answer 2

If test.sh is a bash script, another way could be to export a function named nproc:

bash-5.0$ cat foo.sh
#! /bin/bash

echo make -j $(nproc)
bash-5.0$ ./foo.sh
make -j 8

Creating and exporting a function:

bash-5.0$ nproc() { echo function 10; }
bash-5.0$ (export -f nproc; ./foo.sh)
make -j function 10