Question

**Add comments to existing ARM code to explain steps**

Question that my code answers: Write an ARM assembly program to calculate the value of the following function:

f(y) = 3y^2 – 2y + 10 when y = 3.

My Code below, that needs comments added:

FunSolve.s

LDR R6,=y

LDR R1,[R6]

MOV R2,#5

MOV R3,#6

MUL R4,R1,R1

MUL R4,R4,R2

MUL R5,R1,R3

SUB R4,R4,R5

ADD R4,R4,#8

st B st

y DCD 3

Answer 1

Your code and its meaning is as follows

LDR R6,=y            // Load the address for the value of 'y' as R6

LDR R1,[R6]         // Load the value at the address found in R6 to register R1 (R1 = 3)

MOV R2,#5         // Move the number 5 to R2 (R2 = 5)

MOV R3,#6         // Move the number 6 to R3 (R3 = 6)

MUL R4,R1,R1   // R4 = R1 * R1 (R4 = 3 * 3 = 9)

MUL R4,R4,R2   // R4 = R4 * R2 ( R4 = 9 * 5 = 45)

MUL R5,R1,R3   // R5 = R1 * R3 (R5 = 3 * 6 = 18)

SUB R4,R4,R5   // R4 = R4 - R5 (R4 = 45 - 18 = 27)

ADD R4,R4,#8   // R4 = R4 +8 (R4 = 27 + 8 = 35)

st B st                  // ***********I didn't get it **********

y DCD 3              // Defining a constant data 3 to 'y' ( y = 3)

The constant values you entered in this code will not produce the correct result.

The function given here is

f(y) = 3y^2 - 2y + 10

and then putting y=3

f(3) = 3 * (3 * 3) - 2 * 3 +10

      = 31

The correct code is

LDR R6,=y            // Load the address for the value of 'y' as R6

LDR R1,[R6]         // Load the value at the address found in R6 to register R1 (R1 = 3)

MOV R2,#3         // Move the number 3 to R2 (R2 = 3)

MOV R3,#2        // Move the number 2 to R3 (R3 = 2)

MUL R4,R1,R1   // R4 = R1 * R1 (R4 = 3 * 3 = 9; this indicate y^2)

MUL R4,R4,R2   // R4 = R4 * R2 ( R4 = 9 * 3 = 27; this indicate 3y^2)

MUL R5,R1,R3   // R5 = R1 * R3 (R5 = 3 * 2 = 6; this indicate 2y)

SUB R4,R4,R5   // R4 = R4 - R5 (R4 = 27 - 6 = 21; this indicate 3y^2-2y)

ADD R4,R4,#10 // R4 = R4 + 10 (R4 = 21 + 10 = 31)

STR R1, R4          // Storing the result 31 in register R1 (R1 = 31)

y DCD 3              // Defining a constant data 3 to 'y' ( y = 3)