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From: Kyohei Kadota <lufia@lufia.org>
Date: Tue,  2 Apr 2019 23:44:18 +0900
Message-ID: <CAFMepc=qmOLe_W=-VBe+HCwc+NMrcD7vQoLbtr4xytrn3BW6ag@mail.gmail.com>
To: Fans of the OS Plan 9 from Bell Labs <9fans@9fans.net>
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Subject: Re: [9fans] Don't Plan 9 C compiler initialize the rest of member
 of a struct?
Topicbox-Message-UUID: f81aa6b4-ead9-11e9-9d60-3106f5b1d025

I ran this code on 386 machine on QEMU based public VPC service.

2019=E5=B9=B44=E6=9C=882=E6=97=A5(=E7=81=AB) 9:27 Skip Tavakkolian <skip.ta=
vakkolian@gmail.com>:
>
> It should initialize to zero. 8c and 5c both do the right thing here.
>
> Which distribution and cputype?
>
> On Mon, Apr 1, 2019, 8:34 AM Kyohei Kadota <lufia@lufia.org> wrote:
>>
>> Hi, 9fans. I use 9legacy.
>>
>> About below program, I expected that flags field will initialize to
>> zero but the value of flags was a garbage, ex, "f8f7".
>> Is this expected?
>>
>> ```
>> #include <stdio.h>
>>
>> struct option {
>>     int n;
>>     char *s;
>>     int flags;
>> };
>>
>> int
>> main(void)
>> {
>>     struct option opt =3D {1, "test"};
>>     printf("%d %s %x\n", opt.n, opt.s, opt.flags);
>>     return 0;
>> }
>> ```
>>