blog:2023:08:22:2023-08-22_-_interesting_bash_feature [BaSzErr]

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====== 2023-08-22 - interesting bash feature ======

talk is cheap, so i'll show you some code instead:
<code bash>
#!/bin/bash
set -e
false && echo "this never shows up"
echo "result: $?"
</code>
what will it show? ''set -e'' is there so nothing, right? well -- no. it will print this:

   result: 1

...and returns exit code ''0''. now the open question is: how on earth?! return from the ''false && whatever'' is clearly non-zero, yet the next line gets executed! stating that previous line ended with an error.

let's now compare it with almost identical piece of code:
<code bash>
#!/bin/bash
set -e
( false && echo "this never shows up" )
echo "result: $?"
</code>

the output? nothing + return code of ''1''. just as expected!

spoiler alert -- it's not a bug. it's a feature. just very non-obvious one. [[https://stackoverflow.com/questions/25794905/why-does-set-e-true-false-true-not-exit|here's a nice explanation]]. TL;DR version is that ''set -e'' exits with an error on non-handled error. eg. this would not fail the script:
<code bash>
if [ 1 -eq 2 ]
then
  echo "wtf?!"
fi
</code>
despite ''[ 1 -eq 2 ]'' is clearly false. now if we look at ''foo && bar'' through the same lenses, here ''foo'''s failure is considered "handled", as there's a post-action ''&& bar'', thus ''$?'' is set as expected, but script does not stop.

i found this one while inspecting other's person script. it wasn't really clear what's the problem. in the future -- when you write a shell script, do use ''set -e'' but try keeping 1 statement per line. it makes everyone's lifes easier. :)

thx [[https://pragmatic-architect.com/|Wojtek]] for helping out with this one! :)

blog/2023/08/22/2023-08-22_-_interesting_bash_feature.txt · Last modified: 2023/08/22 19:36 by basz

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