(1)
Convert the ten decimal numbers of List 1 to standardised floating point form, giving the result as the number pair (mantissa, exponent), where
mantissa = sign bit.octal fraction
exponent = decimal integer
For example, 4.5 = (0.44, 1)
(2)
Store the first five numbers of List 1 in standardised floating point form in full words iA1, i = 0 to 4, using only B-type instructions.
(3)
Write in octal form the numbers stored in full words A1 to A10 by the section of ABL program shown in List 2. For example, the contents of A6 are J4, i.e. (A6) = J4.
Exhibit the contents of the accumulator after the execution of each of the following sets of instructions. Use the notation
ay' = J (3 octal digits)
m' = octal digit . (13 octal digits, omitting right hand zeroes)
l' = . (13 octal digits, omitting right hand zeroes)
sign l' = 0 or 1
For example, the instruction
1) + 16
324 0 0 A1
leaves the accumulator in the state
ay' = J002
m' = 0.2
l' = .0
sign l' = 0
(1)
1) -16
324 0 0 A1
(2)
1) +K21.5:13
334 0 0 A1
(3)
1) +K275.123456 +0:26
324 0 0 A1
320 0 0 1A1
(4)
1) +K73:3
334 0 0 A1
340 0 0 J4
(5)
1) -10
344 0 0 A1
101 124 0 A1
163 124 0 0
In problems (6) onward, in addition to showing the final state of the accumulator, indicate accumulator overflow (AO), exponent overflow (SO) and exponent underflow (EU) and instruction(s) which caused them.
(6)
1) -1
346 0 0 J4
331 0 0 A1
(7)
1) -1(:127)
344 0 0 A1
121 124 0 0
355 0 0 J4
(8)
1) +K2,-K3
324 0 0 A1
352 0 0 1A1
(9)
1) -K1(:127)
325 0 0 A1
(10)
1) +5
2) +1(:-24)
324 0 0 A1
320 0 0 A2
(11)
1) +K.4
344 0 0 A1
364 0 0 J4
365 0 0 J4
(12)
1) -1(:64)
314 0 0 A1
372 0 0 A1
365 0 0 J4
(13)
1) +.125(:-127), +0
324 0 0 A1
332 0 0 1A1
366 0 0 J4
(14)
In this problem exhibit the contents of the store line before and after the instruction is performed.
1) -1
335 0 0 A1
356 0 0 A1
(15)
1) +1
2) -1
324 0 0 A1
373 0 0 A2
In questions 1), 2) and 3) the standardised floating point numbers a0, a1, ..., a99 are held in addresses iA79, i = 0 to 99.
(1) Place their double length floating point sun in the accumulator, round to single length and store the result in A80.
(2) Place the integral part of a0 in the accumulator in double length fixed point form (exponent = 25, mantissa = int. pt. a0 × 8-25)
(3) Assume that the accumulator holds a double length floating point number Z. Compute the product of that number, double length, with V = (A90) and leave the result in the accumulator.
In questions 4), 5), 6) and 7) the integers x0, x1, ..., x99 are stored in (iA100), i = 0 to 99 in the substandard floating point form (12, xi × 8-12).
(4) Working fixed point, store their alternating sum x0 - x1 + x2 - ... in A99.
(5) Place the integral part of the quotient x0/x1 in A399 and the remainder in 1A399, both in the substandard form. Assume x0 and x1 to be positive.
(6) Store x3 × x4 as a standardised double length number in A341, 1A341.
(7) Working fixed point, form the quotient 936.29 / 48.01 to 3 octal places and store it in A309 in the substandard form (10, x). Store the remainder in standardised floating point form in 1A309. The operands should be stored by program.
(1)
(2)
(1)
121 1 0 J4
113 1 0 A1
113 0 0 0.4A1
(2)
121 1 0 J007634
113 1 0 1A1
113 0 0 1.4A1
(3)
121 1 0 1U9
113 1 0 2A1
113 0 0 2.4A1
(4)
121 1 0 J773
113 1 0 3A1
113 0 0 3.4A1
(5)
121 1 0 K3U9
113 1 0 4A1
113 0 0 4.4A1
(3)
(A1) = J0061 (A2) = J002765 (A3) = J01377777 (A4) = J064 (A5) = J005777 (A7) = J006144 (A8) = J021777774224 (A9) = J7764 (A10) = J0116
(1)
ay' = J002
m' = 7.6
l' = .0
sign l' = 0
(2)
ay' = J015
m' = 0.00...021 (13 octal place fraction)
l' = .0
sign l' = 0
(3)
ay' = J003
m' = 0.275 (integral part of (A1))
l' = .0
sign l' = 0
(4)
ay' = J002
m' = 0.73
l' = .0
sign l' = 0
(5)
ay' = J001
m' = m
l' = .66
sign l' = 1
(6)
ay' = J0 A0 set by 331
m' = 1.0
l' = .0
sign l' = 0
(7)
ay' = J765
m' = 7.0
l' = .0
sign l' = 0
(8)
ay' = J002
m' = 7.72
l' = .0
sign l' = 1
(9)
ay' = J200 B0 set by 325
m' = 0.1
l' = .0
sign l' = 0
(10)
ay' = J001
m' = 0.5000000000001
l' = .000000000001
sign l' = 0
(11)
ay' = 0
m' = 7.4
l' = .0
sign l' = 0
(12)
372 0 0 A1 gives EO and AO
ay' = J200
m' = 0.1
l' = .0
sign l' = 0
(13)
ay' = J601
m' = .1
l' = .0
sign l' = 0
(14)
The contents of the store line before and after the instruction are the same, i.e. J001 and the 335 instructions sets AO.
(15)
ay' = J001
m' = 0.1
l' = .0
sign l' = 0
(1)
(2)
1) 0.26
324 0 0 A79
330 0 0 A1
364 0 0 J4
122 124 0 J001 to get exponent 25
(3)
Let Z1 and Z2 denote the more and less significant halves of Z, and W1 and W2 the halves of the product VZ. Thus VZ = W1 + W2. Working space is A1, A2...
(4)
(5)
(6)
(7)