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FATEK FBs-CB2
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Advanced Function Instruction
7-112
FUN129 P
MBROT
MATRIX BIT ROTATE
FUN129 P
MBROT
Ms : Starting register of source matrix
Md : Starting register of destination matrix
L : Length of matrix (Ms and Md)
Ms, Md may combine with V, Z, P0~P9 to
serve indirect address application
WX WY WM WS TMR CTR HR IR OR SR ROR DR K XR
Range
Ope-
rand
WX0
WX240
WY0
WY240
WM0
WM1896
WS0
WS984
T0
T255
C0
C255
R0
R3839
R3840
R3903
R3904
R3967
R3968
R4167
R5000
R8071
D0
D4095
2
256
VZ
P0~P9
Ms
Md * *
L *
L
Ms Md
Rotate
left
1 bit
OTB
L/R=1
z When rotate control "EN" = 1 or "EN" ( P instruction)
has a transition from 0 to 1, matrix Ms will be
completely retrieved and rotated by one bit towards
the left (when L/R = 1) or to the right (when L/R = 0).
The space created by the rotation (with a left rotation it
will be M0, and with a right rotation it will be M
16L-1
) will
be replaced by the status of the rotated-out bit (with a
left rotation it will be M
16L-1
, and with a right rotation it
will be M0). The rotated-out bit will not only be used to
fill the above-mentioned space, it will also be
transferred to rotated-out bit "OTB".
L
Ms Md
Shift
right
1 bit
OTB
L/R=0
X0
EN
L :
Md :
5
OTBMs :
R 0
R 0
L/R
129P.MBROT
z In the program at left, Ms and Md are the same matrix.
When X0 goes from 0
1, then the whole of Ms is
retrieved and rotated right (because L/R = 0) by 1 bit. It is
then stored back into Ms itself (because in this example
Ms and Md are the same matrix). The results are shown
at right in the diagram below.
Ms15
Ms
Ms0
Md15
Md
Md0
R0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 R0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
R1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 R1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
R2 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 R2 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0
R3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 R3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
R4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 R4 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Ms79
Ms64
X0=
Ö
Md79
Md64
Before execution After execution
OT
B
0

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