Write a Pep/9 assembler program that divides a number by a number by shifting ri

Question

Write a Pep/9 assembler program that divides a number by a number by shifting right and subtracting. Each shift right halves the number. For example, if we shift 0x02 one bit to the right the answer is 0x01 (0000 0010 shifted left is 0000 0001). Divide 0x28 by 0x10 (in decimal it is 40 divided 16, but leave the answer in hex). Write a Pep/9 assembler program that divides a number by a number by shifting right and subtracting. Each shift right halves the number. For example, if we shift 0x02 one bit to the right the answer is 0x01 (0000 0010 shifted left is 0000 0001). Divide 0x28 by 0x10 (in decimal it is 40 divided 16, but leave the answer in hex).

Explanation / Answer

Consider for above program

(3005H)=0X28

(3006H)=0X10

Program is

LXI 3005H // load Immediate location

MOV A,M // mov memory M to accumulator A

RRC A //Rotate right to A

MOV M,A // Move A to M =>3007H Location

SUB M //substract =>3008H Location

SUB M //3009H Location

SUB M //3010H Location

SUB M //3011H Location

STA 3012H // Store result to 3012H

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