Hi.
I am writing a (32-bit)program in which a very time-critical task is to get
the position of the least significant bit of a long long (64-bit) integer,
and clearing it. I have the following assembler macro utilizing the bsf
instruction to find the lsb :
#define get_lsb_and_clear(__input_bb,__output_sq) \
asm volatile("xorl %%edx, %%edx;"\
"xorl %0, %0;"\
"incl %%edx;"\
"bsfl (%%ebx),%%ecx;"\ //Check for lsb in lower 32-bit dword
"jnz 1f;"\
"leal 4(%%ebx),%%ebx;"\ //If not found, Step to the upper 32-bit dword
"bsfl (%%ebx),%%ecx;"\ //Check for lsb in upper 32-bit dword
"xorl $32,%0;"\
"1: shll %%cl,%%edx;"\
"xorl %%edx,(%%ebx);"\
"xorl %%ecx,%0;": "=&r" (__output_sq):"b"
(&__input_bb):"%edx","%ecx","memory")
This works, but is rather slow. As I understand it, having "memory" in your
clobber-list is bad for performance as the compiler cannot make assumptions
as to the state of the memory over the asm, and since I am interested in
high performance I would like to give the compiler a good chance at
optimization. I have tried to remove this constraint by using "=m"
(__input_bb) as output instead (as I've seen done in some tutorials), but
then the address %1 cannot be changed to the upper 32-bits. How do I get
around this problem? Is there any way of changing the address when using
"=m"? Can I get rid of the "memory" at all?
Thanks for any help.
Rikard Ojala
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