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From: Masahiro Yamada <masahiroy@kernel.org>
To: Linux Kernel Mailing List <linux-kernel@vger.kernel.org>,
	linux-api@vger.kernel.org
Subject: Why is the bit size different between a syscall and its wrapper?
Date: Fri, 12 Mar 2021 11:48:11 +0900	[thread overview]
Message-ID: <CAK7LNAQYMKqixm8dVbbMvgt+=MEROSg-JG_kHS8T+xmrgeLABg@mail.gmail.com> (raw)

Hi.

I think I am missing something, but
is there any particular reason to
use a different bit size between
a syscall and its userspace wrapper?



For example, for the unshare syscall,

unshare(2) says the parameter is int.


SYNOPSIS
       #define _GNU_SOURCE
       #include <sched.h>

       int unshare(int flags);




In the kernel, it is unsigned long.


SYSCALL_DEFINE1(unshare, unsigned long, unshare_flags)
{
        return ksys_unshare(unshare_flags);
}




I guess the upper 32-bit will be
zeroed out in the c library when
sizeof(int) != sizeof(unsigned long)
(i.e. 64-bit system), but I'd like to know
why we do it this way.


-- 
Best Regards
Masahiro Yamada

             reply	other threads:[~2021-03-12  2:50 UTC|newest]

Thread overview: 4+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2021-03-12  2:48 Masahiro Yamada [this message]
2021-03-12  3:17 ` Why is the bit size different between a syscall and its wrapper? Bhaskar Chowdhury
2021-03-12  3:27 ` Willy Tarreau
2021-03-14  5:10   ` Masahiro Yamada

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