On 2007-03-23, Oleg Verych <[email protected]> wrote:
>> From: Grant Edwards
>> Newsgroups: gmane.comp.hardware.texas-instruments.msp430.gcc.user
>> Subject: Question about libgcc 16x16=>32 multiply
>> Date: Thu, 22 Mar 2007 20:23:28 +0000 (UTC)
>
> Hallo.
>
>> uint32_t r;
>>
>> void foo(uint16_t u1, uint16_t u2)
>> {
>> r = (uint32_t)u1 * (uint32_t)u2;
>> }
>>
>>
>> I've been looking at the code for the case above:
>>
>> 1 foo:
>> 2 push r11
>> 3 push r10
>> 4 mov r15, r10
>> 5 mov r14, r12
>> 6 clr r11
>> 7 clr r13
>> 8 call #__umulhisi3
>> 9 mov r14, &_r
>> 10 mov r15, &_r+2
>> 11 pop r10
>> 12 pop r11
>> 13 ret
>>
>>
>> I've noticed that even though a 16x16 multiply is being called,
>
> As i'm new on all this, i just note, that function names are very
> strange.
Huh?
>> the two operands are being passed as 32-bit values.
>
> And this is, what your C expression tells, no?
Yes, but the compiler is smart enough that it has recognized
that it needs to do a 16x16->32 multiply, so it's calling a
16x16->32 multiply:
"umul" means unsigned multiply.
"hi" means 16 bit operands
"si" means 32 bit result
"3" means result doesn't overwrite either operand
>> Why pass two 32 bit values to a library function who's purpose is to
>> multiply two 16 bit values? IOW why are lines 6 and 7 there?
>
> file gcc/config/msp430/libgcc.S:
> ...
> .global __umulhisi3
> .func __umulhisi3
> __umulhisi3:
> br #__mulsi3
> ...
> /*******************************************************
> Multiplication 32 x 32
> *******************************************************/
> ...
> .global __mulsi3
> .func __mulsi3
>
> __mulsi3:
>
> Thus, it seems 32x32 instead.
That's because nobody has bothered to write a 16x16->32
multiply routine. For other targets where that's the case,
there either isn't a umulhisi3 function (the compiler does the
16->32 conversion before generating the instuction), or
umulhisi3 expects 16-bit operands and converts them to 32 bit
values before jumping to the 32x32 multiply.
My question is: why does umulhisi3 expect to be passed 32-bit
operands when it's defined as a 16x16 multiply?
--
Grant Edwards grante Yow! I have no actual
at hairline...
visi.com
