1. The content of AC in the basic computer is hexadecimal 789A ; the initial val
ID: 3550697 • Letter: 1
Question
1. The content of AC in the basic computer is hexadecimal 789A ; the initial
value of E is 1; and the initial value of PC is hexadecimal 048.
Determine the contents of E, AC, PC, AR, and IR in hexadecimal in each of the
four independent register reference instructions below. Thus each instruction
starts with the given Initial values.
E AC PC AR IR
Initial 1 789A 048 - -
CLA ? ? ? ? ?
CLE
SPA
SZA
2. (16 points) An instruction at address 047 in the basic computer has I = 0, and an
address part equal to 025 (all numbers are hexadecimal). The memory word at
address 025 contains 5E34 and the content of AC is 789A.
Determine the contents of registers PC, AR, AC, and IR for the two independent
memory-reference instructions below. Each instruction starts with the given Initial
values.
PC AR AC IR
Initial 047 789A
AND ? ? ? ?
BUN
2
3. (24 points) The following program is a list of instructions for the Basic Computer in
hexadecimal code. The computer executes the instructions starting from address
100.
Show the Symbolic instructions with the AC values in the sequence that they are
executed.
Location Instruction
100 7800
101 2109
102 4107
103 7008
104 4106
105 7200
106 7001
107 7020
108 4103
109 8000
Note:
1) The complement of a hexadecimal character is computed such that the hexadecimal
character plus its complement = 15; i.e. the complement of 0 is F and of A is 5.
2) Do not include instructions that are not executed
Location Symbolic instructions comment AC
? ? ? ? ?
4. (22 points) List the locations and hexadecimal code of the Assembly Language
program below.
The locations should be in hexadecimal starting with 100 and ending with 10B.
Location Hexadecimal Assembly Language
instruction code
ORG 100
100 7800 CLA
_______ ________ CLE
_______ ________ STA MM1
_______ ________ LDA MM2
_______ ________ SZA
_______ ________ BUN MM3
_______ ________ BUN MM4
_______ ________ MM3, CIL
_______ ________ CMA
_______ _______ MM4, HLT
_______ _______ MM1, HEX 0
10B _______ MM2, HEX 62C1
END
5. (4 points) List and briefly define four different kinds of assembly language
statements.
The content of AC in the basic computer is hexadecimal 789A; the initial value of E is 1; and the initial value of PC is hexadecimal 048. Determine the contents of E, AC, PC, AR, and IR in hexadecimal in each of the four independent register - reference instructions below. Thus each instruction starts with the given Initial values. An instruction at address 047 in the basic computer has 1 = 0. and an address part equal to 025 (all numbers are hexadecimal). The memory word at address 025 contains 5E34 and the content of AC is 789A. Determine the contents of registers PC, AR, AC, and IR for the two independent memory - reference instructions below. Each instruction starts with the given Initial values. The following program is a list of instructions for the Basic Computer in hexadecimal code. The computer executes the instructions starting from address 100. Show the Symbolic instructions with the AC values in the sequence that they are executed. Note: The complement of a hexadecimal character is computed such that the hexadecimal character plus its complement = 15; i.e. the complement of O is F and of A is 5. Do not include instructions that are not executed List the locations and hexadecimal code of the Assembly Language program below. The locations should be in hexadecimal starting with 100 and ending with WB. List and briefly define four different kinds of assembly language statements.Explanation / Answer
1.
The corrected table with comments is below:
2.
The corrected table with comments is below:
3.
The table filled in the sequence of execution is as follows:
Location
Symbolic instruction
comment
AC
100
7800 - CLA
Clear the accumulator, AC = 0
0
101
2109 - LDA 109
Load the AC with value at 109
8000
102
4107 - BUN 107
Brach to 107
8000
107
7020 - INC
Increment AC
8001
108
4103 - BUN 103
Brach to 103
8001
103
7008 - SNA
Skip if AC is negative and AC is negative and skips the next instruction
8001
105
7200 - CMA
Compliment AC
7FFE
106
7001 - HLT
Halt the program
7FFE
4)
The Hexadecimal instruction of the give assembly is as follows:
Location
Assembly language
Instruction code in HEXADECIMAL
100
CLA
7800
101
CLE
7400
102
STA MM1
310A
103
LDA MM2
210B
104
SZA
7004
105
BUN MM3
4107
106
BUN MM4
4109
107
MM3 CIL
7040
108
CMA
7200
109
MM4, HLT
7001
10A
MM1 HEX 0
0
10B
MM2 HEX 62C1
62C1
5)
a) CLA - this is an assebly instruction to tell the processor to clear the contents of the Accumulator. Hence Accumulator is loaded with 0x0000.
b) INC - This assembly instruction Increments the contents in the accummulator by 1 and loads the carry(if any) to E.
c) BUN m - This instruction tells the processor to branch unconditionally to the location specified by m.
d) HLT - This instruction is for the processor to HALT the excution of all the instructions.
Hope this helps. Please feel free to contact if any further info is needed.
E AC PC AR IR Comment Initial 1 789A 48 CLA 1 0 49 800 7800 Clears AC CLE 0 789A 49 400 7400 Clears E SPA 1 789A 4A 10 7010 Skip the next instruction if AC is positive. The value in AC is positive so the next instruction is skipped and PC reaches 4A not 49 SZA 1 789A 49 4 7004 Skip the next instruction if AC is zero. The value in AC is not zero so the next instruction is not skipped and PC reaches 49 withput skipping.Related Questions
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