Re: How to get the application path

[Markus E Leypold <development-2006-8ecbb5cc8aREMOVETHIS@ANDTHATm-e-leypold.de>]

> On Jul 23, 5:34 am, Stephen Leake <stephen_le...@stephe-leake.org>
> wrote:
>> Rangdalf <rangd...@hotmail.com> writes:
>> > Hi,
>>
>> > In C# there is something like Application.startupPath. Is there a
>> > package or function to get this path.
>> > I can get the current directory but I need the application one.
>>
>> In Ada 95, Ada.Command_Line.Command_Name returns the executable name,
>> which usually (not always) has the full path
>
> I really have to question this.  I tried it using GNAT on Linux, and
> (as I expected) Command_Name simply returns whatever was the first
> token I entered on the shell command line---i.e. argv[0] (which is
> what the Ada 95 manual hints that Command_Name should return).
>
> Maybe things are different on Windows, but the original poster didn't
> refer to any particular operating system.
>
> ; that is the application
>> path.
>>
>> There may be a better solution in Ada 2005.
>
> If we assume that Command_Name returns the name of some (executable)
> file that may or may not be a full path name, then

Command_Name returns the first element of argv[]. There is a
convention that this is the name of the executable. But since argv[0]
is set by the invoking process, it might be anything: The full path of
an executable, just the executable name (which it usually is on Unix
with the Bourne shell when path searching is in effect) or _anything_
else.

BTW: Is it possible the OP doesn't want the path to the executable,
but rather the directory which becomes /became the wurrent directory
of the application at startup? Isn't that what Application.startupPath
is on Windows?


> Ada.Directories.Full_Name (Ada.Command_Line.Command_Name) should
> return the full path, if the default directory hasn't been changed
> with Ada.Directories.Set_Directory.  To extract just the directory
> name, without the base name, Ada.Directories.Containing_Directory
> (Ada.Directories.Full_Name (Ada.Command_Line.Command_Name)) should
> work.


Regards -- Markus