This is zsh -f:
Fereidoons-MacBook-Pro% local a=$(jaja) && echo bad
zsh: command not found: jaja
bad
Fereidoons-MacBook-Pro% a=$(jaja) && echo bad
zsh: command not found: jaja
Fereidoons-MacBook-Pro%
Why is local messing up error handling?
Add a comment
|
1 Answer
From the zsh manual regarding the typeset builtin (which local is a special case of):
Unlike parameter assignment statements,
typeset's exit status on an assignment that involves a command substitution does not reflect the exit status of the command substitution. Therefore, to test for an error in a command substitution, separate the declaration of the parameter from its initialization:# WRONG typeset var1=$(exit 1) || echo "Trouble with var1" # RIGHT typeset var1 && var1=$(exit 1) || echo "Trouble with var1"
In your case:
$ unset a
$ local a=$(jaja) && echo bad
zsh: command not found: jaja
bad
$ unset a
$ local a && a=$(jaja) && echo bad
zsh: command not found: jaja
