Re: A83: HL\2
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Re: A83: HL\2
At 06:39 1998-08-31 +0900, you wrote:
>
>> Shift h right one step, then rotate l through carry to the right.
>>
>> sra h
>> rr l
>
>Woah, could you explain that a little. So the sra rotates h right one
>step...what does the rr l do? Thanks.
look at it like this:
contents in HL before:
H: L:
10010101 01011100
sra shifts H right one step like this:
a zero is put in here ->01001010 <-the bit that was "pushed out" is put in
Carry (important)
rr rotates L right _through_carry_ like this:
Here is the carry-bit put in (the bit that was pushed out from H)->10101110
th bit pushed out is put in carry here to.. but we wont use that..
so after, HL will look like this:
H: L:
01001010 10101110
thus shifted right one step, which is equal to have divided HL by 2..
I hope that helped...
//Olle
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