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R data
=
00000000
Constant
=
10000000
----.-.:----.,------.-.---......---------
ANDed result
=
00000000
On the branch line, R, you see the
two mnemonics AC and lBC.
AC,ALU carry,
brings up control lines to test for a
carryout condition of ALU as a result of
the arithmetic expression executed in
the previous ROS word
~ddress
03A4,
expression RtD+C->RC).
lBC, one bit
carry, brings up the control lines to
test for a carry into the highest posi-
tion of ALU as a result of the previous
arithmetic statement.
To show the posi-
tions of ALU effected, ASSUME this data:
A-register
=
0100
B-register
=
1100
0000
0000
-------------------~---------
11 carries
ALU output
=
0000
t t
11
If
II
. 11.---1BC
I
I
L--AC
0000
Let·s continue with our example.
The
previous expression RtD+C->RC added the
last two bytes of data from both reg-
isters.
Both bytes of data were zero,
therefore the output from ALU was zero.
fie had neither an ALU carry nor a 1-bit
carry.
A 0,0 branch is taken to address
01D8 which is in Figure 3-26, location
G3.
The AC and lBC mnemonics test to
determine overflow conditions.
An over-
flow condition would have caused branch-
ing to either address 01D9 or 01DA.
ADDRESS 01D8:
The expression *XBB->N
LS
addreSses aX addressable byte of local
storage.
See Figure 3-24.
This byte
contains the condition codes and program
mask bits.
Certain bits are set accord-
ing to whether the answer is equal to,
greater, or less than zero.
Program masks are checked further in
the program.
3-34
A X-addressable byte sets the N-
register to the format:
NO
"
N1
0,
N2
N3
N4
CNO, XO, 1,
N5
N6
N7
Xl, X2, X3.
The
NO
and Nl positions are set
1,
a
respectively •. The 2 position of the
N-register is set by the CN ROS control
field, 0
~it
position.
Position N3 is
set by the CX ROS control field 0
position.
N4 is set to a
1
uncondi-
tionally.
N5,N6, and N7 are set by the
remaining positions of the CX field.
You can see by looking at Figure 3-24
that the coordinates BB address byte 27.
BB in binary is:
B
1011
B
1011
Match this up with the address format
and you see that the CX field must be
coded 1011.
This is the value that
appears on the
E
line of the CAS block.
Format =
1 .. 0,
CNO, XO, 1, Xl, X2, X3,
Value .BB
=
1 0
1 1
1 0 1
1
The control line expression o->SO
sets SO to zero in case we had been in a
complement operation •
The branch tests are S2 and Z
=
O.
S2 was set to a
1
because we had signi-
ficant data.
The expression Z
=
0
checks the Z bus for a zero as a result
of the previous expression on the arith-
metic line (ReXH->Z).
Had the resultant
answer been minus, the expression
ReKH->Zwould have provided a 1-bit
output for the highest position of ALU.
Because our answer is poSitive, the Z
bus is zero and therefore, a 1,1 branch
is executed to address 01EB.
Bad our
answer been minus, the Z= 0 expression
would have set X7 to a O.
A 1,0 branch
would have taken us to address OlEA.
ADDRESS 01EB:
The byte that was just
read from core contains information
pertaining to condition codes and pro-
gram masks.
The high 4-bit positions
are for the condition code settings
according to the answer of the problem.
The expression RL +KH->R presents the
data previously read to the A-register.
A constant (X) sets the B-register to
the value of 2 as specified by line E.
The low portion (L) of the data in the
A-register is used as the A source for
ALU.
The high position
(H)
of the B-
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