8086 program to find Square Root of a number Last Updated : 22 May, 2018 Improve Improve Like Article Like Save Share Report Problem – Write an assembly language program in 8086 microprocessor to find square root of a number. Example – Algorithm – Move the input data in register AX Move the data 0000 in CX and FFFF in BX Add 0002 to the contents of BX Increment the content of CX by 1 Subtract the contents of AX and BX If Zero Flag(ZF) is not set go to step 3 else go to step 7 Store the data from CX to offset 600 Stop Program – OFFSET MNEMONICS COMMENT 0400 MOV AX, [500] AX <- [500] 0404 MOV CX, 0000 CX <- 0000 0407 MOV BX, FFFF BX <- FFFF 040A ADD BX, 02 BX = BX + 02 040E INC CX C = C + 1 040F SUB AX, BX AX = AX – BX 0411 JNZ 040A JUMP to 040A if ZF = 0 0413 MOV [600], CX [600] <- CX 0417 HLT Stop Explanation – M0V AX, [500] is used to move the data from offset 500 to register AX MOV CX 0000 is used to move 0000 to register CX MOV BX FFFF is used to move FFFF to register BX ADD BX, 02 is used to add BX and 02 INC CX is used to increment the content of CX by 1 SUB AX, BX is used to subtract contents of AX with BX JNZ 040A is used to jump to address 040A if zero flag(ZF) is 0 MOV [600], CX is used to store the contents of CX to offset 600 HLT is used end the program Like Article Suggest improvement Previous 8086 program to find the factorial of a number Share your thoughts in the comments Add Your Comment Please Login to comment...