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The content of AC in the basic computer is hexadecimal A937 and the initial value of E is 1. Determine the contents of AC, E, PC, AR, and IR in hexadecimal after the execution of the CLA instruction. Repeat 11 more times, starting from each one of the register-reference instructions. The initial value of PC is hexadecimal 021.?
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The content of AC in the basic computer is hexadecimal A937 and the in...
Solution:

Given: AC = A937, E = 1, PC = 021
CLA Instruction: Clear Accumulator

1. CLA Instruction:
After the execution of CLA instruction, the contents of AC becomes 0000, and E remains the same as 1. The contents of PC, AR, and IR also remain the same.

AC = 0000, E = 1, PC = 021, AR = 0, IR = 0

2. LDA Instruction:
LDA Instruction stands for "Load Accumulator with Memory".
It loads the contents of memory into the accumulator.

- LDA 300: Load the contents of memory location 300 into the accumulator.
After the execution of LDA 300 instruction, the contents of AC becomes the value stored in memory location 300, and E remains the same as 1. The contents of PC, AR, and IR also change.

AC = value at memory location 300, E = 1, PC = next instruction address, AR = 300, IR = LDA 300

Repeat the above steps for the remaining register-reference instructions.

3. STA Instruction:
STA Instruction stands for "Store Accumulator in Memory".
It stores the contents of the accumulator into the memory.

- STA 300: Store the contents of the accumulator into memory location 300.
After the execution of STA 300 instruction, the contents of AC remains the same as the previous value, and E remains the same as 1. The contents of PC, AR, and IR also change.

AC = previous value, E = 1, PC = next instruction address, AR = 300, IR = STA 300

4. ADD Instruction:
ADD Instruction stands for "Add Memory to Accumulator".
It adds the contents of memory to the accumulator.

- ADD 300: Add the contents of memory location 300 to the accumulator.
After the execution of ADD 300 instruction, the contents of AC becomes the sum of the previous value and the value stored in memory location 300, and E remains the same as 1. The contents of PC, AR, and IR also change.

AC = previous value + value at memory location 300, E = 1, PC = next instruction address, AR = 300, IR = ADD 300

5. SUB Instruction:
SUB Instruction stands for "Subtract Memory from Accumulator".
It subtracts the contents of memory from the accumulator.

- SUB 300: Subtract the contents of memory location 300 from the accumulator.
After the execution of SUB 300 instruction, the contents of AC becomes the difference between the previous value and the value stored in memory location 300, and E remains the same as 1. The contents of PC, AR, and IR also change.

AC = previous value - value at memory location 300, E = 1, PC = next instruction address, AR = 300, IR = SUB 300

6. JMP Instruction:
JMP Instruction stands for "Jump to Memory Location".
It changes the value of PC to the specified memory location.

- JMP 300: Jump to memory location 300.
After the execution of JMP 300 instruction, the contents of PC becomes 300, and the contents of AC and E remain the same as the previous value. The contents of AR and IR also change.

AC = previous value, E = 1, PC = 300, AR
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The content of AC in the basic computer is hexadecimal A937 and the initial value of E is 1. Determine the contents of AC, E, PC, AR, and IR in hexadecimal after the execution of the CLA instruction. Repeat 11 more times, starting from each one of the register-reference instructions. The initial value of PC is hexadecimal 021.?
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The content of AC in the basic computer is hexadecimal A937 and the initial value of E is 1. Determine the contents of AC, E, PC, AR, and IR in hexadecimal after the execution of the CLA instruction. Repeat 11 more times, starting from each one of the register-reference instructions. The initial value of PC is hexadecimal 021.? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The content of AC in the basic computer is hexadecimal A937 and the initial value of E is 1. Determine the contents of AC, E, PC, AR, and IR in hexadecimal after the execution of the CLA instruction. Repeat 11 more times, starting from each one of the register-reference instructions. The initial value of PC is hexadecimal 021.? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The content of AC in the basic computer is hexadecimal A937 and the initial value of E is 1. Determine the contents of AC, E, PC, AR, and IR in hexadecimal after the execution of the CLA instruction. Repeat 11 more times, starting from each one of the register-reference instructions. The initial value of PC is hexadecimal 021.?.
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