Moikka! Here's Adrián. I work for Igalia.

On shell builtins

Today I will explain a «misfeature» of the Bash shell which may puzzle you at first sight, but which has its own internal logic. Take the following example:

f () {
  echo "I return false"
  return 1

g1 () {
  local v=$(f)
  echo "Exit status is $?"

g2 () {
  local v
  echo "Exit status is $?"


This script would produce the following output:

Exit status is 0
Exit status is 1

You may wonder why the call to g1 does not print 1. The reason is that local is no more than a shell built-in command: first, the command expansion executes and the exit status variable $? is set to 1, but then the local built-in executes and succeeds, and the exit status variable gets set to zero before we inspect it!

The bottom line is: never ever expand a command in the same line where you declare variables.