[A83] Re: HL = A * 12 (fast)

To: <assembly-83@lists.ticalc.org>
Subject: [A83] Re: HL = A * 12 (fast)
From: "Gijs Leegwater" <g_leegwater@hotmail.com>
Date: Wed, 24 Apr 2002 16:49:34 +0200
References: <3CC6443C.950976B8@dingoblue.net.au> <009a01c1eac8$beec4ac0$d905040a@tijl> <3CC764C5.52D90AD0@dingoblue.net.au> <20020424115719.381BA18062@mx-1.sollentuna.net> <03a301c1eb8a$7ad18450$d905040a@tijl>
Reply-To: assembly-83@lists.ticalc.org


I'd like to say something about it, because when i was a beginning with ASM
i was always confused about it.
Maybe you knwo already but the bitshifting is not a processor operation, the
assembler just shifts the constant $1234 8 bits to the right so the hex code
doesnt even include the byte $34

Bye,
Cheiz
>
> > A little question that can be a little bit outer the subject..
> > Can somone explain to me what mul12table >> 8 mean ?
>
> In TASM at least, '>>' is the right shift operator. The 8 means it's
shifted
> 8 bits. mul12table is address and shifting it 8 bits gives you the higher
> byte.
>
> 1234h >> 8 = 12h
>
>
>

References:

[A83] HL = A * 12 (fast)

From: Darren <qarnos@dingoblue.net.au>

[A83] Re: HL = A * 12 (fast)

From: "Tijl Coosemans" <tijl@ulyssis.org>

[A83] Re: HL = A * 12 (fast)

From: Darren <qarnos@dingoblue.net.au>

[A83] Re: HL = A * 12 (fast)

From: Xavier LaRue <paxl@videotron.ca>

[A83] Re: HL = A * 12 (fast)

From: "Tijl Coosemans" <tijl@ulyssis.org>