Variable expansion is different in zsh from that i

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.

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}

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.