The following is a simple test case for what I want to illustrate.
In bash,
# define the function f
f () { ls $args; }
# Runs the command `ls`
f
# Runs the fommand `ls -a`
args="-a"
f
# Runs the command `ls -a -l`
args="-a -l"
f
But in zsh
# define the function f
f () { ls $args }
# Runs the command `ls`
f
# Runs the fommand `ls -a`
args="-a"
f
# I expect it to run `ls -a -l`, instead it gives me an error
args="-a -l"
f
The last line in the zsh on above, gives me the following error
ls: invalid option -- ' '
Try `ls --help' for more information.
I think zsh is executing
ls "-a -l"
which is when I get the same error. So, how do I get bash's behavior here?
I'm not sure if I'm clear, let me know if there is something you want to know.