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SubjectRe: [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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