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RE: behaviour of bash with "--login" + "--rcfile" depends on position of argument

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  RE: behaviour of bash with "--login" + "--rcfile" depends on position of argument "Göbbert, Jens Henrik" <j.goebbert@fz-juelich.de> - 2019-10-21 14:08 +0000

#15528 — RE: behaviour of bash with "--login" + "--rcfile" depends on position of argument

Thank you, Greg, for the good explanation and your time
and sorry for the wrong bug-report.

Best,
Jens Henrik

________________________________________
From: Greg Wooledge [wooledg@eeg.ccf.org]
Sent: Monday, October 21, 2019 3:43 PM
To: Göbbert, Jens Henrik
Cc: bug-bash@gnu.org
Subject: Re: behaviour of bash with "--login" + "--rcfile" depends on position of argument

On Mon, Oct 21, 2019 at 09:19:48AM +0000, "Göbbert, Jens Henrik" wrote:
> Description:
>         "bash --rcfile --login test.sh" will run test.sh but "bash --login --rcfile test.sh" will not, even though this would be expected from the help/man pages.

--rcfile takes an argument.  It has to be followed by the name of the
rc file that you want bash to read.

When you write   --rcfile --login   you're telling bash that its rc file
is named "--login", and that it should read this file, instead of ~/.bashrc,
if it needs to read an rc file.

However, since you also gave the name of a script to execute, bash doesn't
need to read an rc file.  So, it simply ignores the   --rcfile --login
that you gave it, and just executes test.sh.

On the other hand, when you write   --login --rcfile test.sh
you are telling bash that its rc file is named "test.sh", and that it
should read this instead of ~/.bashrc if it needs to read an rc file.

But since you're using the   --login   argument, bash doesn't need to
read an rc file.  So, it doesn't make use of the   --rcfile test.sh
arguments.  You're basically just running   bash --login   .


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