Worked example 1

Trace the following assembly program. Assume memory is initially all zeros. What is the final value stored in memory location 502?

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// Program Code
LDM #10     // Load the number 10 into the ACC
ADD #5      // Add the number 5 to the ACC
STO 502     // Store the ACC's content at address 502
END

Trace Table:

Step	Instruction	ACC before	ACC after	Memory Changed
1	LDM #10	0	10	-
---	---	---	---	---
2	ADD #5	10	15	-
3	STO 502	15	15	M[502] = 15

Working:

LDM #10: Immediate addressing. The value 10 is loaded directly into the Accumulator (ACC). ACC is now 10.
ADD #5: Immediate addressing. The value 5 is added to the current content of the ACC (10). ACC becomes 10 + 5 = 15.
STO 502: Direct addressing. The content of the ACC (15) is stored in the memory location with address 502.

Final Answer: The final value stored in memory location 502 is 15.

Worked example 2

A program needs to find the largest number in a list of three numbers stored at memory addresses 200, 201, and 202. The values are: M[200]=7, M[201]=12, M[202]=5. Trace the program and state the final value in the Accumulator.

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// Program Code
LDX #0        // Initialise Index Register to 0
LDR 200,X     // Load first number into ACC
LOOP: INC X   // Increment Index Register
      CMP 200,X // Compare ACC with next number
      JPE ENDLOOP // If equal, no change needed
      JPN ENDLOOP // If ACC is greater, no change
      LDR 200,X // If new number is greater, load it
ENDLOOP: CMP X,#2 // Have we checked all numbers?
         JPN LOOP  // If IX < 2, loop again
END

Initial State: ACC=0, IX=0, M[200]=7, M[201]=12, M[202]=5

Trace Table:

Instruction	IX before	ACC before	Comparison	IX after	ACC after	Notes
LDX #0	0	0	-	0	0	IX = 0
---	---	---	---	---	---	---
LDR 200,X	0	0	-	0	7	ACC = M[200+0] = 7
LOOP 1
INC X	0	7	-	1	7	IX is now 1
---	---	---	---	---	---	---
CMP 200,X	1	7	7 vs 12	1	7	Compare ACC(7) with M201
JPE ENDLOOP	1	7	False	1	7	Not equal, continue
JPN ENDLOOP	1	7	False	1	7	ACC is not greater, continue
LDR 200,X	1	7	-	1	12	New number is larger, ACC = M[201] = 12
CMP X,#2	1	12	1 vs 2	1	12	IX < 2, so loop again
JPN LOOP	1	12	True	1	12	Jump to LOOP
LOOP 2
INC X	1	12	-	2	12	IX is now 2
---	---	---	---	---	---	---
CMP 200,X	2	12	12 vs 5	2	12	Compare ACC(12) with M202
JPE ENDLOOP	2	12	False	2	12	Not equal, continue
JPN ENDLOOP	2	12	True	2	12	ACC is greater, jump to ENDLOOP
CMP X,#2	2	12	2 vs 2	2	12	IX is not less than 2
JPN LOOP	2	12	False	2	12	Do not jump, continue to END

Final Answer: The program terminates with the value 12 in the Accumulator.

Assembly Language — practice questions

Worked example 1

Worked example 2