Question

Accumulator A starts with a value of 200d in decimal, Accumulator B starts with a value of 0AH, which should be the final values accumulator A and B end up with after executing the MUL AB instruction on the 8051 architecture.

A. A = 07H, B = D0H
B. B = 07H, A = D0H
C. A = 24H , R0 = 24H
D. A = 23H , R0 = 23H

Answer 1

The final values of Accumulator A and R0 after executing the MUL AB instruction on the 8051 architecture.

Step-by-step explanation:

The MUL AB instruction on the 8051 architecture multiplies the values in Accumulator A and Accumulator B and stores the result in Accumulator A and the higher 8 bits of the result in the R0 register. To understand the final values after executing this instruction, we need to convert the initial values to binary and perform the multiplication.

1. Accumulator A starts with a value of 200d, which is equivalent to 11001000b.
2. Accumulator B starts with a value of 0AH, which is equivalent to 00001010b.
3. Multiplying these binary numbers, we get the result as 0110011000b, which is equal to 198d.
4. Therefore, the final values will be: Accumulator A = 198d, R0 = 1100b (12d).