Date: Mon, 11 Aug 2003 02:04:10 +0100
From: Jamie Lokier <jamie@shareable.org>
To: linux-kernel@vger.kernel.org
Subject: Re: NULL.  Again.  (was Re: [PATCH] 2.4.22pre10: {,un}likely_p())
Message-ID: <20030811010410.GC10446@mail.jlokier.co.uk>
References: <1060087479.796.50.camel@cube> <20030809002117.GB26375@mail.jlokier.co.uk> <20030809081346.GC29616@alpha.home.local> <20030809015142.56190015.davem@redhat.com> <1060425774.4933.73.camel@dhcp22.swansea.linux.org.uk> <20030809162332.GB29647@mail.jlokier.co.uk> <20030809173001.GG24349@perlsupport.com> <6ufzkad8w1.fsf@zork.zork.net>
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Sean Neakums wrote:
> I had thought that the need for writing NULL where a pointer is
> expected in varags functions was because the machine may have
> different sizes for pointers and int.  In the case of the second
> printf call above, if pointers are 64-bit and integers are 32-bit,
> printf will read eight bytes from the stack, and only four will have
> been occupied by the integer 0.

It is incorrect to write NULL as a varargs argument.  It won't
necessarily pass a pointer because:

   1. a valid definition of NULL is "0".

If you want to pass a pointer to varargs, you might have considered
writing "(void *) 0", but that doesn't work because:

   2. a machine may have different sizes for different pointer types.

You must write "(type) 0" or "(type) NULL", using the correct pointer type.

-- Jamie
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