zsh is not splitting by IFS after parameter expansion

user_1175080 0
Viewed 1039

This is my code to loop over colon separated values and do something with each value.

f()
{
    IFS=:
    for arg in $1
    do
        echo arg: $arg
    done
}

f foo:bar:baz

This works fine in most POSIX compliant shells.

$ dash foo.sh
arg: foo
arg: bar
arg: baz
$ bash foo.sh
arg: foo
arg: bar
arg: baz
$ ksh foo.sh
arg: foo
arg: bar
arg: baz
$ posh foo.sh
arg: foo
arg: bar
arg: baz
$ yash foo.sh
arg: foo
arg: bar
arg: baz

But it does not work as expected in zsh.

$ zsh foo.sh
arg: foo:bar:baz

Is zsh in violation of POSIX here?

ifs zsh posix shell bash
2 Answers

Yes. Zsh has chosen its own way.

Here is the zsh faq entry: “3.1: Why does $var where var="foo bar" not do what I expect?”

In this particular case, you could workaround by adding the -y option to the zsh invocation:

$ zsh -y foo.sh
arg: foo
arg: bar
arg: baz

You could take a look at the zsh's faq especially the chapter 2 and 3. The more you've experienced other shells, the more you can find zsh's pitfalls.

user_43870390

In Zsh it is usually cleaner to split with the provided (s) flag (vs using IFS). A solution for your data would then be:

% f() { for e in ${(s.:.)1}; print $e }
% f foo:bar:baz
foo
bar
baz

See the PARAMETER EXPANSION section in zshexpn(1) man page for more details and related flags.

(I assume you mean colon-separated values.)

user_3265160
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