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ANSWERS -- SYSTEM/360 INSTRUCTIONS
1.
Halfwords
2.
Two, Specification
3.
Op Code
4.
a.
1A
00011010
1
b.
56
01010110
2
c.
9C
10011100
2
d.
FD
11111101
3
5.
Displacelf,ent, Base Address
6.
Index
1.
Does not
8.
Base Registers
9.
4095
10.
1 to 15
11.
The contents of Register 0 are
ignored and a value of zero is used
for the Base or Index factor.
12.
r-----· -T----'
a.
RR
lOp CodelR1 R21
L _____ --1-____
J
r-~---T-----~--,
b.
RX
lOp CodelRl X2
1
1
B2\ 021
l _ _
---.L _ _
--1--~
__
J
r-------~--~~--l
c.
RS
lOp CodelR1 R31B21D21
l ______
~
_____
~
__ J
r-------------T--T--'
d.
SI
lOp Code
12
IB11D11
l ____________
~_~J
r-----~--T--~-T--T-~T--l
e.
5S
lOp CodeJL 1 1L21B11Dl1B2JD21
l _______
~_4-_~~
__
~_~_J
13.
1st
14.
7
APPENDIX 3 •• ANSWERS TO REVIEW QUESTIONS
15.
The Instruction, Byte
16.
RX
11.
SS
18.
The number of Bytes in the data
field is one greater than the num-
ber in the length code.
ANSWERS -- INSTRUCTION SEQUENCING AND
BRANCHING
1.
Program Status Word
2.
64
3.
40
-
63
4.
2, 4,
6
5.
0-' ; Op Code
6.
·Current·
1.
Some type of internal register
storage area
8.
Instruction address,
(location)
9.
34 - 35
10.
4
11.
Some
12.
Condition Code
13.
<zero
00
01
10
11
14.
Equal or Low
15.
Always
address
or
16.
Bits
8 - 31
of general register 5
17.
The effective address generated by
adding the contents of register 4
and register 7 and a displacement
factor of O.
6-19
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