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Subject: Re: [Bug] Exiting shell from function called by trap handler always
 produces status 0
To: zsh-workers@zsh.org
References: <CGME20181009012624epcas1p44f2ae223f663713a980af4be735e5a3f@epcas1p4.samsung.com>
 <a5934536-7982-c434-d98d-2c07d66b0739@inlv.org>
 <20181009084918eucas1p27dedda10d51beb773ba9175967912d2c~b5IUtvcVK3037630376eucas1p2R@eucas1p2.samsung.com>
From: Martijn Dekker <martijn@inlv.org>
Message-ID: <205ea5e4-4dc9-2ebf-9437-0822195e1a9a@inlv.org>
Date: Tue, 9 Oct 2018 12:46:20 +0100
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Op 09-10-18 om 09:49 schreef Peter Stephenson:
> On Mon, 2018-10-08 at 14:02 +0100, Martijn Dekker wrote:
>> When a trap handler exits the shell using the 'exit' command within a
>> function, the shell's exit status is zero even if another exit status
>> was given as an argument to the 'exit' command.
>>   
>> $ Src/zsh -c 'fn() { exit 13; }; trap "fn" EXIT'
>> $ echo $?
>> 0
>> (expected output: 13)
> diff --git a/Src/builtin.c b/Src/builtin.c
> index c5b319b..b81acdb 100644
[...]

This still doesn't fix it. Output identical to above.

> +  $ZTST_testdir/../Src/zsh -fc 'fn() { exit 13; }; trap fn EXIT; exit'

The extra '; exit' at the end makes it work correctly, but should not be 
necessary.

Thanks,

- M.