But as we know, the kerword 'register' is only a suggestion. The
language specification does not force to put the 'register' variable
into a register.
Putting the variable onto stack does not violate anything, but the
assembler will fail.
I think the hack does not really work. THX.
2007/7/26, H. Peter Anvin <[email protected]>:
> Jian-Xin Lai wrote:
> > Hi,
> >
> > The kernel version is 2.6.20.7. In include/asm-i386/string.h, line 169~185:
> > static inline char * strrchr(const char * s, int c)
> > {
> > int d0, d1;
> > register char * __res;
> > __asm__ __volatile__(
> > "movb %%al,%%ah\n"
> > "1:\tlodsb\n\t"
> > "cmpb %%ah,%%al\n\t"
> > "jne 2f\n\t"
> > "leal -1(%%esi),%0\n" (*)
> > "2:\ttestb %%al,%%al\n\t"
> > "jne 1b"
> > :"=g" (__res), "=&S" (d0), "=&a" (d1) (**)
> > :"0" (0),"1" (s),"2" (c)
> > :"memory");
> > return __res;
> > }
> >
> > The 'lea' instruction needs a register here (*), but "g" (**) means
> > any registers, memory or immediate integer. Is it correct? If the
> > compiler do not place the __res into a register, the compilation will
> > fail. Should it be "r"?
> > Thank you very much.
> >
>
> Yes, it should be "r", and it looks like the author hacked around the
> fact they had the wrong constraints by using the "register" keyword on
> the variable declaration.
>
> -hpa
>
--
Regards,
Lai Jian-Xin
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