Question

Write a Hack Assembly Language program to calculate the quotient from a division operation. The values of dividend a and divisor b are stored in RAM[0] (R0) and RAM[1] (R1), respectively. The dividend a is a non-negative integer, and the divisor b is a positive integer. Store the quotient in RAM[2] (R2). Ignore the remainder.

Example: if you are given two numbers 15 and 4 as dividend a (R0) and divisor b (R1), respectively, the answer will be 3 stored in R2.

test cases:

| RAM[0] | RAM[1] | RAM[2] |

| 1 | 1 | 1 |

| 2 | 2 | 1 |

| 0 | 2 | 0 |

Answer #1

// Storing R0 value into a new variable a

R0

X = Y // X= R0

a

Y = X // a = R0

// Storing R1 value into a new variable b

R1

X = Y // X = R1

b

Y= X // b = R1

// Initializing R2 as 0

0

X = Z // X = 0

R2

Y = X // R2 = 0

(NOTZERO)

a

X = Y // X = a

b

X = X - Y // X = a - b

END

X;JLT // (a-b) < 0 go to END

ZERO

X;JEQ // If (a - b == 0) go to ZERO

R2

Y = Y + 1 // R2 = R2 + 1

b

X = Y // X = b

a

Y = Y - X // a = a - b

NOTZERO // Going back to start of loop

0;JMP

(ZERO)

R2

Y = Y + 1 // R2 = R2 + 1

(END)

END

0;JMP

VII. After execution of the following instructions what value
will be in register r2? Trace the program and write the value of
the registers at each line to receive full credit.
LDR r12, =0xA4000000
LDR r0, =0x2D
STR r0, [r12, #-4]!
LDR r0, =0x5E
STR r0, [r12, #-4]!
LDR r0, =0xD5
STR r0, [r12, #-4]!
LDMIA r12!, {r0-r2}
SUB r2, #0x68
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