Re: A short but unsuccessful script for calling a Java app.

On Fri, 26 Aug 2005 21:38:09 UTC, "Monique Y. Mudama"
<spam@xxxxxxxxxxxxxxxx> opined:
> On 2005-08-26, Stan Goodman penned:
> >
> > I'm sure that the directory is not in the path, because I just made
> > it. I am surprised that it has to be in the path when I am running
> > the file from within the directory; I am coming from OS/2, where
> > running a file from within its directory doesn't require a path
> > entry (because there is nothing to seek). I'll get used to it. See
> > below.
>
> This is actually a security feature of sorts. What if you want to see
> the contents of the current directory, and the current directory has
> an executable named 'ls' that actually removes your home directory, or
> similar?
>
> If you really want the current directory to be in the path, you just
> need to add . to the $PATH variable. But in practice all you need to
> do is type ./my_executable instead of my_executable, which is only two
> more letters and keeps you from accidentally running the wrong
> executable. Well, sort of. It can't help it if you have two entries
> in your PATH, both of which contain executables of the same name.
>

As you point out toward the end, one can't anticipate every possible
weird eventuality, and there is such a thing as running into
diminishing returns if one tries too hard to do so.

But it doesn't matter. I obviously accept that this is the convention,
and I will get used to it.
--
Stan Goodman
Qiryat Tiv'on
Israel

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