What part of this one-line script is specific to zsh?

Question

I wrote a simple program called sarcasm that simply takes all its arguments, concatenates them, and converts the resulting string use 'aLtErNaTiNg CaPs'. I have written a simple one-line shell script that will let me use dmenu to enter text, run that text through this program, and copy the result to my clipboard. Here is what I have:

echo "dmenu -p 'Text: '" | sh | read var ; [[ -n "$var" ]] && echo -n $var | xargs sarcasm | xclip -selection clipboard

It is supposed to:

Prompt me for text using dmenu
If (and only if) I entered any text, run it through the sarcasm program and copy the result to my clipboard

It works fine when I run that exact command with zsh, or if I run it in a script using zsh (i.e. the shebang is #!/bin/zsh), but when I run it with bash, it doesn't copy anything to my clipboard. What part of this is zsh-only, and what is the bash equivalent (if there is one)?

It's all POSIX shell (except for [[ .. ]] which is irrelevant) - nothing is zsh specific. I suspect your ENVIRONMENT is different when you run from bash and it does not have the X environment present (this is a strong suspicion, but cannot be verified without seeing your environment) — David C. Rankin
– David C. Rankin, Commented Jan 14, 2021 at 3:19
@DavidC.Rankin [[ is not POSIX, but it is a legal bash operator. — Shane Bishop
– Shane Bishop, Commented Jan 14, 2021 at 3:21
We might need to see the contents of sarcasm to identify the issue. Or it could be something with your environment. Have you tried echo -n 'test' | xclip -selection clipboard with an interactive bash shell, to try to rule things out? — Shane Bishop
– Shane Bishop, Commented Jan 14, 2021 at 3:23
For instance, echo $var may produce slightly different results. Say var contains the text A B (4 embedded spaces) , then it would produce A B under zsh, but A B (1 space) under bash. — user1934428
– user1934428, Commented Jan 14, 2021 at 8:05

glenn jackman · Accepted Answer · 2021-01-14 03:42:41Z

6

This part:

echo "dmenu -p 'Text: '" | sh | read var 
# ............................^^^^^^^^^^

In bash, each command in a pipeline is run in a separate subshell. Because the var variable is set in a subshell, the variable vanishes when the subshell exits.

To execute that in bash, redirect the output of a process substitution: this way, the read command runs in the current shell.

read -r var < <(dmenu -p 'Text: ')

I imagine that would work in zsh too.

There is another workaround: set the lastpipe setting and disable job control

$ echo foo | read var; declare -p var
bash: declare: var: not found

$ set +o monitor
$ shopt -s lastpipe
$ echo foo | read var; declare -p var
declare -- var="foo"

answered Jan 14, 2021 at 3:42

glenn jackman

249k42 gold badges233 silver badges362 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

AJ-Williams1 Over a year ago

Your first suggestion, redirecting the output of process substitution, worked! Thanks so much!

Collectives™ on Stack Overflow

What part of this one-line script is specific to zsh?

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest