Question

Consider the following instruction sequence where register R1, R2 and R3 are general purpose and MEMORY[X] denotes the content at the memory location X.

Instruction	Semantics	Instruction Size (bytes)
MOV R1, (5000)	R1 ← MEMORY[5000]	4
MOV R2, (R3)	R2 ← MEMORY[R3]	4
ADD R2, R1	R2 ← R1 + R2	2
MOV (R3), R2	MEMORY[R3] ← R2	4
INC R3	R3 ← R3 + 1	2
DEC R1	R1 ← R1 – 1	2
BNZ 1004	Branch if not zero to the given absolute address	2
HALT	Stop	1

 

Assume that the content of the memory location 5000 is 10, and the content of the register R3 is 3000. The content of each of the memory locations from 3000 to 3010 is 50. The instruction sequence starts from the memory location 1000. All the numbers are in decimal format. Assume that the memory is byte addressable.

After the execution of the program, the content of memory location 3010 is ______

Answer 1

Answer: 50 to 50 

Explanation: 

Consider following memory cells

Address	Value
5000	10
3000	50
3001	50
3002	50
3003	50
3004	50
3005	50
3006	50
3007	50
3008	50
3009	50
3010	50

After Executing till 3 instruction we have the following value in Registers

R1	R2	R3
10	50	3000

After 4th instruction, M[R₃] ( M[3000]) will be updated as 50 + 10 = 60 

Address	Value
5000	10
3000	60
3001	50
3002	50
3003	50
3004	50
3005	50
3006	50
3007	50
3008	50
3009	50
3010	50

R₃ => R₃ +1 => 3001

R₁ = 9 Hence it is Nonzero; It will Branch to 1004 (which is instruction 2)

- R₂ = 50

R₂= R₁+R₂ = 59

M[R3] ( M[3001]) will be updated as 59.

Address	Value
5000	10
3000	60
3001	59
3002	50
3003	50
3004	50
3005	50
3006	50
3007	50
3008	50
3009	50
3010	50

R₁  = R₁ - 1 => 9-1 =>8 Hence it is Nonzero; It will Branch to 1004 (which is instruction 1)

If you try to observe the Pattern

Here M[3001] is updated as 59

M[3002] will be updated as 58

M[3003] will be updated as 57

M[3004] will be updated as 56

M[3005] will be updated as 55

M[3006] will be updated as 54

M[3007] will be updated as 53

M[3008] will be updated as 52

M[3009] will be updated as 51

R1 = 0

M[3010] will be updated as 50