How to launch a huge number of processes in bash

Question

I am trying to run a.out lot of times from command line, but am not able to start the processes in background because bash treats it as a syntax error.

for f in `seq 20`; do ./a.out&; done //incorrect syntax for bash near '&'

How can I place & on command line so that bash doesn't complain, and I am allowed to run these processes in background, so that I can generate load on the system.

P.S: I don't want to break it into multiple lines.

Answer 1

Break that into multiple lines or remove the ; after the &

Answer 2

& is a command terminator as well as ; ; do not use both.

And use bash syntax instead of using seq, which is not available on all Unix systems.

for f in {1..20} ; do ./a.out& done

Answer 3

This works:

for f in `seq 20`; do ./a.out& done

& terminates a command just like ; or &&, ||, |.

This means that bash expects a command between & and ; but can't find one. Hence the error.

Answer 4

Remove the ; after a.out:

for f in `seq 20`; do ./a.out& done