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Date: Tue, 9 Jun 2009 00:36:48 +0200
Message-ID: <e01d8a50906081536i34f7710bu49588c85e37d2a13@mail.gmail.com>
Subject: Re: Bug in $PATH initialization?
From: Lennart Borgman <lennart DOT borgman AT gmail DOT com>
To: cygwin AT cygwin DOT com
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On Tue, Jun 9, 2009 at 12:28 AM, ABCD<en DOT ABCD AT gmail DOT com> wrote:
> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> Lennart Borgman wrote:
>> I am trying to get "bash -i" to start up with $PATH having
>>
>> =C2=A0 =C2=A0/usr/local/bin:/usr/bin:/bin
>>
>> at the beginning. I believe I need to do this without any init file
>> running so I set the windows Path variable to contain the
>> corresponding windows directories first. So I set Path in cmd.exe with
>>
>> =C2=A0 set Path=3Dc:\cygwin\bin;%Path%
>> =C2=A0 set Path=3Dc:\cygwin\usr\bin;%Path%
>> =C2=A0 set Path=3Dc:\cygwin\usr\local\bin;%Path%
>>
>> and then I do
>>
>> =C2=A0 bash -i
>>
>> However $PATH starts with
>>
>> =C2=A0 =C2=A0/usr/local/bin:/usr/bin:/usr/bin
>>
>> I thought the last dir above should be /bin, not /usr/bin.
>>
>> Is this intentional or is it a bug. If it is intentional can I somehow
>> do what I want another way?
>>
>
> In Cygwin, /bin and /usr/bin both always point to the same place, so it
> does get the effect you seem to want, just not in the way you expected.

Thanks, but why do I see both /usr/bin and /bin in a cygwin shell
started with --login then?

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