Variable expansion is different in zsh from that i

Question

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.

Answer 1

The difference is that (by default) zsh does not do word splitting for unquoted parameter expansions.

You can enable “normal” word splitting by setting the SH_WORD_SPLIT option or by using the = flag on an individual expansion:

ls ${=args}

or

setopt SH_WORD_SPLIT
ls $args

If your target shells support arrays (ksh, bash, zsh), then you may be better off using an array:

args=(-a -l)
ls "${args[@]}"

---

From the zsh FAQ:

• 2.1: Differences from sh and ksh

  The classic difference is word splitting, discussed in question 3.1; this catches out very many beginning zsh users.

• 3.1: Why does $var where var="foo bar" not do what I expect? is the FAQ that covers this question.

From the zsh Manual:

• 14.3 Parameter Expansion

  Note in particular the fact that words of unquoted parameters are not automatically split on whitespace unless the option SH_WORD_SPLIT is set; see references to this option below for more details. This is an important difference from other shells.

• SH_WORD_SPLIT

  Causes field splitting to be performed on unquoted parameter expansions.