Reply to message «Check existence of a program»,
sent 21:29:11 01 February 2011, Tuesday
by Anonymous bin Ich:
prog=exiftime
path==$prog
if [[ $? -ne 0 ]] ; then
prog=identify
path==$prog
endif
It works because zsh takes first = as assignment operator and expands second
`=$prog' construct into a full path. If `=$prog' expansion fails, it throws an
exception, exception prevents variable from being set and thus last expression
fails what is indicated in $?.
If you use $(cmd) construct, then though `cmd' fails, $(cmd) that does not care
about exit code just expands into output of `cmd', so last expression (which is
variable assignment, NOT `cmd') does not fail. Also note that `=$prog' does not
produce new fork.
Original message:
> Hello!
>
> I am having trouble checking for existence of a program.
>
> This works:
>
> % cat working.zsh
> #!/bin/zsh
> set -x
> prog="identify"
> path=$(which ${prog})
> %
> % ./working.zsh
> +./working.zsh:3> prog=identify
> +./working.zsh:4> path=+./working.zsh:4> which identify
> +./working.zsh:4> path=/usr/bin/identify
> %
>
> But this doesn't:
>
> % cat notworking.zsh
> #!/bin/zsh
> set -x
> prog="exiftime"
> path=$(which ${prog})
> if [[ ${?} -ne 0 ]]; then
> prog="identify"
> path=$(which ${prog})
> fi
> %
> % ./notworking.zsh
> +./notworking.zsh:3> prog=exiftime
> +./notworking.zsh:4> path=+./notworking.zsh:4> which exiftime
> +./notworking.zsh:4> path='exiftime not found'
> +./notworking.zsh:5> [[ 1 -ne 0 ]]
> +./notworking.zsh:6> prog=identify
> +./notworking.zsh:7> path=+./notworking.zsh:7> which identify
> +./notworking.zsh:7> path='identify not found'
> %
>
> Any idea?
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