Question

Assume that operand 30 is at the memory location 200 , operand 40 is at the next word, and the final operand 45 is at the last word. Find the content of the register R3, R4, R5, R6, R7, R8, R9, and (R9) after executing each code (i.e., each line ) (We assume the word length is 32 bits)

Move R3,#200→R3= ?
Load R4,(R3) →R4= ?
Load R5, 4(R3) →R5=?
Load R6, 8(R3) →R6= ?
Add R7, R5, R4 → R7=?
Add R8, R7, R6 → R8=?
Add R9, R3, #12 → R9=?
Store R8, (R9) → (R9)=?

Answer 1

The question relates to assembly language tasks involving loading and storing data in registers, and performing calculations. Register values are computed by executing one instruction at a time following memory addresses and arithmetic operations as given in the problem statement.

Step-by-step explanation:

The question involves assembly or machine language instructions, specifically about register manipulation and memory addressing. Let's go through each code line and compute the values.

Move R3,#200→ R3 will contain the address, so R3=200.

Load R4,(R3)→ R4 loads the value at the address in R3, hence R4=30.

Load R5, 4(R3)→ Given word length is 32 bits (4 bytes), R5 loads the value at R3+4 bytes, so R5=40.

Load R6, 8(R3) → Now R6 will load the value at R3+8 bytes, which means R6=45.

Add R7, R5, R4 → Addition of R5 and R4 gives R7=70 (40+30).

Add R8, R7, R6 → Adding R7 and R6 gives R8=115 (70+45).

Add R9, R3, #12 → Increment address in R3 by 12, which means R9=212.

Store R8, (R9) → The value in R8 is stored at the address specified by R9, so (R9)=115.