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Re: Return status of command substitution with $(...) "gets lost"
From: |
Marc Herbert |
Subject: |
Re: Return status of command substitution with $(...) "gets lost" |
Date: |
Wed, 10 Mar 2010 17:28:11 +0000 |
User-agent: |
Thunderbird 2.0.0.23 (X11/20090825) |
Chet Ramey a écrit :
> On 3/4/10 2:36 AM, Ettelbrueck, Heiko wrote:
>> Problem: The $? variable is always 0 after that statement. (If,
>> on the other hand, I separate the declaration and the
>> definition of the variable as shown in the example below, the
>> $? variable is really set to the exit status of the external
>> tool.)
>
> The exit status is the status of the command you run: local. In the
> absence of a command, when there are only assignment statements, the
> exit status can be the exit status of a command substitution:
I have been bitten by this a few times. Since then I never, ever write this:
local foo=...
But always:
local foo; foo=...
It's overkill, but simple. I tend to even forget why I am doing this,
which is nice: now I can focus on the real problems to solve.
- Return status of command substitution with $(...) "gets lost", Ettelbrueck, Heiko, 2010/03/04
- Re: Return status of command substitution with $(...) "gets lost", Chris F.A. Johnson, 2010/03/04
- Re: Return status of command substitution with $(...) "gets lost", Chet Ramey, 2010/03/04
- Re: Return status of command substitution with $(...) "gets lost",
Marc Herbert <=
- Re: Return status of command substitution with $(...) "gets lost", Eric Blake, 2010/03/10
- Re: Return status of command substitution with $(...) "gets lost", Chet Ramey, 2010/03/10
- Message not available
- Re: Return status of command substitution with $(...) "gets lost", Chet Ramey, 2010/03/11
- Re: Return status of command substitution with $(...) "gets lost", Marc Herbert, 2010/03/11
- Re: Return status of command substitution with $(...) "gets lost", Eric Blake, 2010/03/11
- Re: Return status of command substitution with $(...) "gets lost", Chet Ramey, 2010/03/11