Re: Can't tell the difference in operation between PATH_SCRIPT and NO_PATH_SCRIPT

[Steve Dondley <s@dondley.com>]

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Sorry, I’ve got a bad misconcepiton somewhere. I stil don’t get it.

I do: PATH=.dir

When I do: setopt pathscript
And then do: foo.zsh

It finds foo in the ./dir directory and executes it.


Then when i do: unsetopt pathscript
And then I do: foo.zsh

It stil finds foo.zsh in the ./dir directory.

So I’m not seeing how it behaves any differently. 





> On Jan 25, 2024, at 10:05 PM, Lawrence Velázquez <larryv@zsh.org> wrote:
> 
> On Thu, Jan 25, 2024, at 9:00 PM, Steve Dondley wrote:
>> I’ve been looking over the documentation and I really can’t make sense 
>> out of what the PATH_SCRIPT option is supposed to do.
>> 
>> If it’s set, which it is by default, I think it’s supposed to look 
>> through directories PATH and try to find the command. But if it’s not 
>> set, it doesn’t look through directories in PATH.
>> 
>> At least that’s my understanding. But as a far as I can tell, I can run 
>> a command in /bin with or without PATH_SCRIPT option set.
>> 
>> I’m obviously confused about what the docs are saying. Can someone shed 
>> light on this for me?
> 
> The option affects how zsh behaves when invoked without -c or -s
> and with a first non-option argument that doesn't contain any
> slashes.  In this situation, shells generally try to find and run
> a script with that name in the current directory; if there is no
> such script, they try to find one in PATH.  PATH_SCRIPT lets you
> control whether zsh performs that fallback PATH search.  (It does
> not affect other PATH searches.)
> 
> 	% cat foo.zsh
> 	cat: foo.zsh: No such file or directory
> 	% cat dir/foo.zsh
> 	print foo
> 	% PATH=./dir
> 	% /bin/zsh -o PATH_SCRIPT foo.zsh
> 	foo
> 	% /bin/zsh +o PATH_SCRIPT foo.zsh
> 	/bin/zsh: can't open input file: foo.zsh
> 
> -- 
> vq


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