Question: Pre-Execution state of Register/Memory Location values: PC: 35 Memory Location Value 35 901 (INPUT) [Assume user inputs 500 from keyboard] 36 181 (ADD 81) 37
Pre-Execution state of Register/Memory Location values:
PC: 35
| Memory Location | Value |
| 35 | 901 (INPUT) [Assume user inputs 500 from keyboard] |
| 36 | 181 (ADD 81) |
| 37 | 383 (STORE 83) |
| 38 | 901 (INPUT) [Assume user inputs 901 from keyboard] |
| 39 | 583 (SUBTRACT 83) |
| 40 | 902 (OUTPUT A) |
| 41 | HALT |
| 81 | 400 |
| 82 | 000 |
| 83 | 000 |
* Note: 901 and 902 are instruction for input and output contents of Accumulator.
The following code is executed. Compute the result. Show the values of the registers in each of the following step. Mark Fetch/Execute Cycles within the code. Write a summary on what this program does.
Instruction: 901 (Input)
PC -> MAR MDR -> IR IOR -> A
PC+1 -> PC
Instruction: ADD 81
PC -> MAR MDR -> IR
IR[address] -> MAR
A+MDR -> A
PC +1 -> PC
Instruction: STORE 83
PC -> MAR MDR -> IR IR[address] -> MAR
A -> MDR PC+1 -> PC
Instruction: 901 (Input)
PC -> MAR MDR -> IR IOR -> A
PC+1 -> PC
Instruction: SUBTRACT 83
PC -> MAR MDR -> IR
IR[address] -> MAR
A-MDR -> A
PC+1 -> PC
Instruction: 902 (OUTPUT A)
PC -> MAR MDR -> IR
A -> IOR
PC+1 -> PC
// This is the halt instruction. Assume the content on that memory location is HALT and the system knows to halt the program when that instruction is loaded.
PC -> MAR MDR -> IR
WHAT IS DISPLAYED AS THE FINAL ANSWER? __________
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