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

[Steve Dondley <s@xxxxxxxxxxx>]

OK, thanks.

At the risk of dragging this on, allow me to ask this simple question to make sure I got it:

Is there any way to run this command by itself on the command line:

> setopt pathscript

And have it affect the outcome of any subsequent commands? It looks like the answer is no. 

It looks to me like the pathscript setting can only affect future commands if it’s either:

a) used with +o or -o as part of an argument passed directly to zsh while running a script

b) or if it’s set in a .zshrc file (or other config file) and then sourced when a new shell (interactive mode)

Please tell me I have this correct otherwise I’m still badly confused. 

On Jan 25, 2024, at 11:58 PM, Lawrence Velázquez <larryv@xxxxxxx> wrote:

On Thu, Jan 25, 2024, at 11:17 PM, Steve Dondley wrote:

I guess I never realized that, when it comes to options, there is a 
difference between running a script directly and passing it as an 
argument to zsh. 

You're overgeneralizing.  PATH_SCRIPT is defined to only affect the
latter scenario, but you are confused because you have been trying
to apply it to the former scenario.  There is no broader lesson
about options here.

It's like being confused that POSIX_CD doesn't make the jobs(1)
builtin more POSIX-conformant, even though it never claimed to.
It is a category mistake.

But even still, If I do this:

echo $PATH

/usr/bin:./dir

ls dir

-rwxr--r-- 1 root root 9 Jan 26 03:26 foo.zsh

setopt pathscript

zsh foo.zsh

zsh: can't open input file: foo.zsh

unsetopt pathscript

zsh foo.zsh

zsh: can't open input file: foo.zsh


So I’m still seeing no difference between execution of the script with 
path script on or off in these cases.

Options are not inherited by child shells created in this manner.
That is why my demonstration used -o and +o.

% ([[ -o EXTENDED_GLOB ]]; print $?)
1
% setopt EXTENDED_GLOB
% ([[ -o EXTENDED_GLOB ]]; print $?)
0
% zsh -c '[[ -o EXTENDED_GLOB ]]; print $?'
1

HOWEVER, I discovered if I put this in my .zshrc:

setopt pathscript

and do:

zsh -i foo.zsh

It works.

Yes, because interactive shells source .zshrc.

Or, if I do

setopt pathscript

and then do

zsh -c foo.zsh

This also works.

No, you are conflating unrelated things again.  With -c, zsh evaluates
the "foo.zsh" argument as a complete script, so it performs a PATH
search as usual.  As I said earlier, PATH_SCRIPT *does not apply* to
''zsh -c''; its status is irrelevant.

% cat foo.sh
cat: foo.sh: No such file or directory
% cat dir/foo.sh
echo foo
% PATH=./dir /bin/zsh -o PATH_SCRIPT -c foo.sh 
foo
% PATH=./dir /bin/zsh +o PATH_SCRIPT -c foo.sh
foo


-- 
vq