Answer:
Following is given the answer step by step:
Explanation:
First of all we will write a program using MARIE that will support the statement: Sum = (A + B) - (C + D). All the necessary comments are given in front of each statement:
Load A # variable A will be loaded
Add B # B will be added to A
Store Temp1 # A + B will be stored in Temp1
Load C # C will be loaded in memory
Add D # D will be added to C
Store Temp2 # C + D will ge stored in Temp2
Load Temp1 # Temp1 will be loaded in the memory
Subt Temp2 # Temp2(A + B) get subtracted from Temp1(A - B)
Store Sum # (A + B) - (C + D) will get stored in Sum.
We can see from above program that memory is accessed 9 times. While if C + D get executed first than memory accesses will be reduced to 7.
Above same program could be written using an architecture of 4 registers:
The program is as follows:
Load R1 , A #A will be loaded into R1
Load R2 , B # B will be loaded into R2
Add R1 , R2 # R2 gets added to R1 and the result is stored in R1 (A + B)
Load R3 , C # C loaded into R3
Load R4 , D
# D loaded into R4
Add R3 , R4 # Value in R4 gets added into R3 and R3 becomes (C + D)
#no memory accesses required for this operation
Subt R1 , R4 #R4 (C + D) gets subtracted from R1 (A + B)
#no memory accesses required for this operation
Store Sum # The recent value will be stored into Sum
Here memory is accessed 5 times in total.
<h2>I hope it will help you!</h2>
