Re: Check existence of a program

[ZyX <zyx.vim@gmail.com>]

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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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