Control Data 3100 Reference Manual page 109

Rev. B
23 18 17 16 1 5 14
12
11 1
b
k
b = index register designator
k = shift count; K = k
+
(B
b
)
00
Instruction Description: Refer to SHA description.
23 18 17 16 15 14
13 1 0 1 b
1
k
b = index register designator
k=shift count; K=k
+
(B
b
)
00
(Approximate execution time: 1.8-3.8
J.Lsec.)
(Approximate execution time: 1.8-3.7
~sec.)
Instruction Description: (B
b
)
and k, with their signs extended, are added. If b=O, the sign of
k is still extended. The sign and magnitude of the 24-bit sum determine the direction and
magnitude of shift. The computer senses only bits 00-05 and 23 of the sum for this informa-
tion. For a left shift, the magnitude is placed in k; to shift right, the complement of the
shift magnitude is placed in k.
Examples: (b
=
0 in both cases):
Shift left three places:
Shift right three places:
k=00003
k=77774
Comments: During left shifts bits reaching the upper bit position of the A register are
carried end around to the lowest bit position of Q. Therefore, a left shift of 48 places re-
sults in no change in (AQ). A left shift exceeding 48 places results in an effective shift of
K-48 places.
During right shifts, the sign bit is extended and the bits are shifted end-oft'. A right shift
of 47 or more places results in (AQ) becoming all "O's" or all "l's", depending upon the
original sign.
23 18 17 16 1 5 14
13 11 1
b
k
b
=
index register designator
K
=
k minus the shift count
K~Bb
00
(Approximate execution time: 2.6-5.1
J.Lsec.)
Instruction Description:
(AQ) is shifted left, end around, until the
2
highest order bits (46 and
47) are unequal.
If
(AQ) should initially equal positive or negative zero, 4810 shifts are exe-
cuted before the instruction terminates. During scaling, the computer counts the number
of shifts. A quantity K, called the residue, is equal to k minus the shift count. If b = 0, this
quantity is discarded; ifb = 1,2, or
3,
the residue is transferred to the designated index register.
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