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To: zsh-users@zsh.org
From: Georg Wittig <nc-wittigge@netcologne.de>
Subject: Global And Local Variables in zsh Functions
Message-ID: <55C68669.8000408@netcologne.de>
Date: Sun, 9 Aug 2015 00:44:57 +0200
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Hi,

I'm writing a zsh script that's constructing a rather complex rsync
command. One part of the script is constructing  a list of exclude
commands for rsync. For this I use a global variable EXC. It is built
in the following way

EXC=''
function my_f1 () {
  EXC+=" --exclude='$1'";
}
my_f1 /tmp
my_f1 /opt/windows7
echo ">>>$EXC<<<"

Function my_f1 does a lot of other things that are not relevant in
this context. The output of that script snippet is

>>> --exclude='/tmp' --exclude='/opt/windows7'<<<

which is exactly what I want.

Now I need that function my_f1 return another value that depends on
it's input parameter. So I rewrote my_f1 to my_f2 which looks as follows

EXC=''
function my_f2 () {
    EXC+=" --exclude='$1'";
    local x="some value depending on $1"
    echo $x
}
x1=$(my_f2 /tmp)
x2=$(my_f2 /opt/windows7)
echo ">>>$EXC<<<"

To my surprise, EXC is empty now: The output is

>>><<<

Why is EXC empty in the case of my_f2, and correct in the case my_f1?
Exists there an influence of the local variable x, or is the culprit
the way that my_f2 returns it's value to the calling point? How do I
rewrite my_f2 such that the value of EXC is correct?

Thanks for your hints,
     --Georg

[zsh-5.0.8-5.fc22.x86_64]