On Mon, Nov 10, 2008 at 4:26 PM, Nick Craig-Wood <[EMAIL PROTECTED]> wrote: > > Looking at the assembler it produces (x86) > > mul: > pushl %ebp > xorl %edx, %edx > movl %esp, %ebp > movl 12(%ebp), %eax > imull 8(%ebp), %eax > popl %ebp > ret > > Which I'm pretty sure is a 32x32->64 bit mul (though my x86 assembler > foo is weak).
My x86 assembler is also weak (or perhaps I should say nonexistent), but I think this does exactly what the C standard says it should: that is, it returns just the low 32-bits of the product. Looking at the assembler, I think the imull does a 32-bit by 32-bit multiply and puts its (truncated) result back into the 32-bit register eax. I'd guess that the 64-bit result is being returned to the calling routine in registers edx (high 32 bits) and eax (low 32 bits); this explains why edx has to be zeroed with the 'xorl' instruction. And if we were really expecting a 64-bit result then there should be an unsigned multiply (mull) there instead of a signed multiply (imull); of course they're the same modulo 2**32, so for a 32-bit result it doesn't matter which is used. Mark _______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
