talk is cheap, so i'll show you some code instead:
#!/bin/bash set -e false && echo "this never shows up" echo "result: $?"
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:
#!/bin/bash set -e ( false && echo "this never shows up" ) echo "result: $?"
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. 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:
if [ 1 -eq 2 ] then echo "wtf?!" fi
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 Wojtek for helping out with this one! :)