csh script variable substitution

Question

I have a small script that performs the following actions. I have simplified it for this example.

set logfile=$HOME/LOG/output.log
find execute.sh -type f -exec csh -c '"$1" >& "$logfile" &' {} \;

The issue is that the find command is not expanding the $logfile value. Instead it is giving me an error

logfile: Undefined variable.

I am not that familiar with csh, nor have I done scripting for quite a while, but is it possible to escape this or otherwise gain access to the value in the command?

Don't use `csh`; read [Csh programming considered harmful](http://www-uxsup.csx.cam.ac.uk/misc/csh.html). Try `zsh`, `fish` or `bash` — Basile Starynkevitch, Sep 27 '13 at 17:03
Thanks. I've read it and made my case in the past but don't have a choice in the matter. — McArthey, Sep 30 '13 at 15:29

score 1 · Accepted Answer · answered Sep 27 '13 at 16:32

logfile is not an environment variable, so it doesn't exist in the shell started by the -exec primary. You can either use setenv:

setenv logfile "$HOME/LOG/output.log"

or change the quotes so that $logfile is expanded before passing the command to csh. It's not clear to me how $1 is supposed to be set, though.

score 0 · Answer 2 · answered Jul 19 '20 at 19:47

[ I recognize how old this question is, but it still shows up early in searches ] In plain C-shell (/bin/csh, and emulators which restrict syntax to match), the variable setting must be delimited with spaces, thus

set logfile = $HOME/LOG/output.log

In Tcsh, the spaces are optional. If your script was written for strict C-shell compatibility, your first line ended up setting a rather peculiar variable to an empty string: on my Mac, it would have been logfile=/Users/kelsey/LOG/output.log = "".

csh script variable substitution

2 Answers2