Hi,
Was just reading up on these two documents:
http://map.grauw.nl/resources/z80instr.php
and
http://www.msx.org/wiki/R800_Programming (Under caption New Commands)
Makes me wonder: Does MULUB A,(HL) exist?
When I follow Grauw's table for the bitfield encoding it should be:
MULUB A,r => ED C1+8*r
where r can be A, B, C, D, E, H, L or (HL).
Register bits value:
A = 7
B = 0
C = 1
D = 2
E = 3
H = 4
L = 5
(HL) = 6
So it should be: ED F1
On the wiki however I only see:
ED F9 MULUB A,A
ED C1 MULUB A,B
ED C9 MULUB A,C
ED D1 MULUB A,D
ED D9 MULUB A,E
ED E1 MULUB A,H
ED E9 MULUB A,L
ED C3 MULUW HL,BC
ED D3 MULUW HL,DE
ED E3 MULUW HL,HL
ED F3 MULUW HL,SP
GNF
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