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Re: Shell script ignores $PATH?!

On  5 Oct, Brian Dessent wrote:
>  If you want to specify what shell is used to run a script you either 
>  need to specify it in the shebang of the script (#!/path/to/shell) or 
>  you need to start that shell explicitly (/path/to/shell 
>  /path/to/script).  If you try to execute a script with no shebang then 
>  the behavoir is going to be system-dependent.  On cygwin that means 
>  defaulting to /bin/sh, as you can see from 
>        if (buf[0] != '#' || buf[1] != '!') 
>          { 
>            pgm = (char *) "/bin/sh"; 
>            arg1 = NULL; 
>          } 

Thanks, Brian.

It'd be nicer if it instead looked for sh in PATH, but the above is
perfectly reasonable.  I think it's entirely fair that the behaviour is
system dependent: I agree that what I was doing was too ambiguous.

Now that I know the cause, I can cope.

BTW, I've wondered for many years why the #! notation doesn't allow a
pathless program name, to mean "look it up in PATH" in the normal way.

The number of times I've changed "#!/usr/local/bin/perl" to
"#!/opt/bin/perl" or "#!/usr/bin/perl" and back again, and for other
interpreters, is quite large.  Seems like an obvious long-standing
missing Unix feature.  (Apologies for the off-topic rant.)


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