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Date:	Sun, 20 Feb 2011 19:08:45 +0100
From:	Borislav Petkov <bp@...en8.de>
To:	Matthew Wilcox <matthew@....cx>
Cc:	Borislav Petkov <bp@...64.org>, Ingo Molnar <mingo@...e.hu>,
	Dan Carpenter <error27@...il.com>,
	"Herrmann3, Andreas" <Andreas.Herrmann3@....com>,
	Thomas Gleixner <tglx@...utronix.de>,
	Ingo Molnar <mingo@...hat.com>,
	"H. Peter Anvin" <hpa@...or.com>,
	"maintainer:X86 ARCHITECTURE..." <x86@...nel.org>,
	"open list:AMD MICROCODE UPD..." <amd64-microcode@...64.org>,
	open list <linux-kernel@...r.kernel.org>,
	"kernel-janitors@...r.kernel.org" <kernel-janitors@...r.kernel.org>
Subject: Re: [patch -next] x86, microcode, AMD: signedness bug in
 generic_load_microcode()

On Sun, Feb 20, 2011 at 10:50:11AM -0700, Matthew Wilcox wrote:
> On Sun, Feb 20, 2011 at 03:14:52PM +0100, Borislav Petkov wrote:
> > int f() {
> >         return 0xa5a5a5a5;
> > }
> > 
> > int main()
> > {
> > 
> >         char ret = f();
> > 
> >         printf("ret = 0x%016x\n", ret);
> > 
> >         return 0;
> > } 
> > --
> > 
> > doesn't cause a warning and prints a sign extended 0x00000000ffffffa5
> > which is cast to the return type of the function. If ret is an unsigned
> > char, then we return a 0x00000000000000a5.
> > 
> > I found something about it in the C99 standard??, section "6.5.16.1 Simple
> > assignment":
> > 
> > 4.  EXAMPLE 1       In the program fragment
> > 
> >            int f(void);
> >            char c;
> >            /* ... */
> >            if ((c = f()) == -1)
> >                     /* ... */
> > 
> > the int value returned by the function may be truncated when stored in
> > the char, and then converted back to int width prior to the comparison.
> > In an implementation in which ??????plain?????? char has the same range
> > of values as unsigned char (and char is narrower than int), the result
> > of the conversion cannot be negative, so the operands of the comparison
> > can never compare equal. Therefore, for full portability, the variable c
> > should be declared as int."
> > 
> > so the whole "... may be truncated.. " could mean a lot of things. From
> > my example above, gcc does truncate the int return type to a byte-sized
> > char only when they differ in signedness.
> 
> No, that's not what's going on.  GCC _is_ truncating to a byte, 0xa5,
> whether it's signed or not.  Then at the time of the call to printf,
> the 0xa5 is cast to int.  If the char is signed, 0xa5 is sign-extended;
> if unsigned, it's zero-extended.

Yes, you're right, I missed the fact that printf does convert its
arguments based on the format string. I should've done

	printf("ret = 0x%hhx\n", ret);

for chars.

Thanks.

-- 
Regards/Gruss,
    Boris.
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