On Wednesday 26 June 2002 05:26, Sridhar J (june end) wrote:
> Doesn't it mean that a.out is in the current directory? So why I should go
> to a parent directory as in ./a.out to execute it?
Sorry, you're not understanding a couple of things. First, "./" is the
*current* directory, not the paren
On Wed, 26 Jun 2002 17:56:41 +0530
"Sridhar J (june end)" <[EMAIL PROTECTED]> wrote:
> Hello
>
> Thanks to all of you who replied promptly to my question. I would like to
> clarify one thing. When I compiled the program using gcc, I tried typing
> a.out. When that didn't work, I did a ls -l whic
Sridhar J (june end) wrote:
> Hello
>
> Thanks to all of you who replied promptly to my question. I would like to
> clarify one thing. When I compiled the program using gcc, I tried typing
> a.out. When that didn't work, I did a ls -l which showed me a file called
> a.out*.
>
> Doesn't it mean t
On Wed, 26 Jun 2002, Sridhar J (june end) wrote:
> Thanks to all of you who replied promptly to my question. I would like to
> clarify one thing. When I compiled the program using gcc, I tried typing
> a.out. When that didn't work, I did a ls -l which showed me a file called
> a.out*.
>
> Doesn'
Hello
Thanks to all of you who replied promptly to my question. I would like to
clarify one thing. When I compiled the program using gcc, I tried typing
a.out. When that didn't work, I did a ls -l which showed me a file called
a.out*.
Doesn't it mean that a.out is in the current directory? So wh
