On Sun, Feb 20, 2011 at 11:42:17AM -0700, Matthew Wilcox wrote:
> On Sun, Feb 20, 2011 at 07:08:45PM +0100, Borislav Petkov wrote:
> > On Sun, Feb 20, 2011 at 10:50:11AM -0700, Matthew Wilcox wrote:
> > > 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);
>
> GCC's special treatment of the printf format string is only in the
> gneration of warnings. It doesn't promote differently based on the
> format string.
>
> You need to look at 6.5.2.2, parts 6 and 7. Part 7 says:
>
> The ellipsis notation in a function prototype declarator causes
> argument type conversion to stop after the last declared
> parameter. The default argument promotions are performed on
> trailing arguments.
>
> And part 6 describes the default argument promotions:
>
> If the expression that denotes the called function has a type that
> does not include a prototype, the integer promotions are performed
> on each argument, and arguments that have type float are promoted
> to double. These are called the default argument promotions.
>
> So passing a char to printf will cause it to be promoted to int, no
> matter what the format string says. All the format string will do is
> change how it's printed. Probably by casting it back to a char :-)
Ha, I see, maybe I should've seen this earlier if I would've looked at
the asm, as grandma always taught me:
char ret = f();
...
printf("ret = 0x%hhx\n", ret);
translates to:
00000000004004e4 <f>:
4004e4: 55 push %rbp
4004e5: 48 89 e5 mov %rsp,%rbp
4004e8: b8 a5 a5 a5 a5 mov $0xa5a5a5a5,%eax
4004ed: c9 leaveq
4004ee: c3 retq
00000000004004ef <main>:
4004ef: 55 push %rbp
4004f0: 48 89 e5 mov %rsp,%rbp
4004f3: 48 83 ec 10 sub $0x10,%rsp
4004f7: b8 00 00 00 00 mov $0x0,%eax
4004fc: e8 e3 ff ff ff callq 4004e4 <f>
400501: 88 45 ff mov %al,-0x1(%rbp)
400504: 0f be 55 ff movsbl -0x1(%rbp),%edx <--- mov 8-bit reg/mem with sign extension to a 32-bit reg
400508: b8 1c 06 40 00 mov $0x40061c,%eax
40050d: 89 d6 mov %edx,%esi
40050f: 48 89 c7 mov %rax,%rdi
400512: b8 00 00 00 00 mov $0x0,%eax
400517: e8 c4 fe ff ff callq 4003e0 <printf@plt>
vs the unsigned char case
unsigned char ret = f();
...
printf("ret = 0x%hhx\n", ret);
=>
...
400501: 88 45 ff mov %al,-0x1(%rbp)
400504: 0f b6 55 ff movzbl -0x1(%rbp),%edx <--- mov 8-bit reg/mem with zero-extension to a 32-bit reg
400508: b8 1c 06 40 00 mov $0x40061c,%eax
40050d: 89 d6 mov %edx,%esi
40050f: 48 89 c7 mov %rax,%rdi
400512: b8 00 00 00 00 mov $0x0,%eax
400517: e8 c4 fe ff ff callq 4003e0 <printf@plt>
Thanks for enlightening me! :)
--
Regards/Gruss,
Boris.
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