Question

I have a little shell script that runs anything executable in a designated directory, while passing along two arguments:

#!/bin/sh

cd '/foo'
for s in bar/*;
do [ -x $s ] && $s "$1" "$2" || : ;
done

Is the || : bit serving any purpose? I wanted the executables to be kicked off asynchronously, so I changed the ; to a &, but is there any reason I shouldn’t just do:

#!/bin/sh

cd '/foo'
for s in bar/*;
do [ -x $s ] && $s "$1" "$2" &
done

Answer 1

|| : (which is often spelled || true) means you don’t care whether the command succeeds or not. It forces a successful return status ($?).

It is good practice to run shell scripts under the shell’s -e option so that errors don’t go unnoticed. This option can be either activated on the shebang line (#!/bin/sh -e) or in the script itself (set -e). It causes errors to abort execution of the script.

Since in this case you are not running under -e, the || : doesn’t make a different because the shell will ignore the result code anyway if the command fails. Furthermore, you are running a command in the background in this case so checking the return code doesn’t even make sense.

In general, including || true even if you are not running under -e can be good for two reasons:

• To make it compatible with -e in case it is ever changed to run under -e.
• To serve as documentation to someone reading the shell script that you are conscious of the possibility that this command can fail and that such a failure is explicitely OK.