Find Square Root of a hexadecimal number in assembly language

To write 8086 Assembly Language Program to Find Square Root of a hexadecimal number.

Answer:

; SQUARE ROOT OF A NUMBER
MODEL SMALL
.STACK 100
.DATA ; Data segment starts
NUM1 DW 0019H ; Initialize num1 to 0019
SQRT DW 01 DUP (?) ; Reserve 1 word of uninitialised data space to offset sqrt
.CODE ; Code segment starts
START:
MOV AX,@DATA ;Initialize data segment
MOV DS,AX
MOV AX,NUM1 ;Move the number(num1) to AX
XOR BX,BX ;XOR is performed and result is stored in BX
MOV BX,0001H ;Initialize BX to 0001H
MOV CX,0001H ;Initialize CX to 0001H
LOOP1: SUB AX,BX ;AX=AX-BX
JZ LOOP2 ; If zero flag is zero jump to loop2
INC CX ; Increment CX by 1
ADD BX,0002H ;BX=BX+0002H
JMP LOOP1 ; Jump to loop1
INC CX ; Increment CX by 1
LOOP2: MOV SQRT,CX ; Store result
INT 03H ; halt
END START




INPUT:
NUM1 = 0019 H (25 IN DECIMAL)
NUM1 = 0090 H (144 IN DECIMAL)
OUTPUT:
RES = 0005H
RES = 000CH


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AIM: To write 8086 Assembly Language Program to Find Number of one’s & zero’s in a hexadecimal number.

Answer:

MODEL SMALL
.STACK 100
.DATA ; Data segment starts
NUM DW 0F0F0H ; Initialize num to F0F0 H
ONES DB ? ; Reserve 1 byte for storing number of zeros
ZEROS DB ? ; Reserve 1 byte for storing number of ones
.CODE ; Code segment starts
START:
MOV AX,@DATA ;Initialize DS
MOV DS,AX
MOV CL,10H ;Initialize CL to 10H
MOV BX,0000H ;Initialize BX to 0000H
MOV AX,NUM ;Move contents of num to AX
UP:
SHL AX,1 ;Shift the contents of AX left once
JC AHEAD ; If CF=1 jump to label ahead
INC BL ; Increment BL by 1
JMP DOWN ; Jump to the label down
AHEAD:
INC BH ; Increment BH by 1
DOWN:
DEC CL ; Decrement CL by 1
JNZ UP ; If zero flag is not 0 jump to label up
MOV ONES,BH ;Move contents of BH to ones
MOV ZEROS,BL ;Move contents of BL to zeros
INT 03H ; Halt
END START




INPUT:
NUM = 0F0F0 H (0000 1111 0000 1111 0000 b)

OUTPUT:
ONES = 08 H
ZEROS = 08 H






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AIM: To write 8086 Assembly Language Program to Find Factorial of a hexadecimal number.

Answer:

; This ALP will find factorial from 01 H to 08 H

MODEL SMALL
.STACK 100
.DATA ; Data segment starts
NUM DW ?
RES DB 1H
FACT DW ?
.CODE ; Code segment starts
START:
MOV AX,@DATA ; Initialize DS
MOV DS,AX
MOV AH,01 ; Move 01 to AH to take input at runtime
INT 21H ; Dos interrupt
AND AX,000FH ; Logical AND of AX,000f and stores in AX
MOV NUM,AL ; Move contents of AL to val1
MOV CX,AL ; Move contents of AL to CL
MOV AX,RES ; Move contents of res(1) to AX
M:
MUL CX ; AX = AX*CX
LOOP M ; Decrements CX by 1 continues multiplication until CX is 0
MOV FACT,AX ; Value of AX is moved to fact
INT 03H ; Halt
END START





INPUT
8 (DURING RUNTIME)
OUTPUT:
FACT = D980 H










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AIM: To write 8086 Assembly Language Program to Convert BCD to 7-Segment display using lookup table.

Answer:

MODEL SMALL
.STACK 100
.DATA ; Data segment starts and defines array A having
; Equivalent 7 segment codes

A DB 3FH, 06H, 5BH, 4FH, 66H, 6DH, 7DH, 07H, 7FH, 6FH

C EQU 06 ; BCD number which is to be converted
7SEG DB ? ; 7 segment code
.CODE ; Code segment starts
START:
MOV AX,@DATA ; Initialize data segment
MOV DS,AX
MOV BX,OFFSET A ; Moves the starting address of A to BX
MOV AL,C ; Move C value to AL register
XLAT ; It i'll add [BX+AL] i.e. 0000 + 06 = 0006 H
MOV 7SEG,AL ; Store the result
INT 03H ; Halt
END START




INPUT:
C = 06 H
OUTPUT:
AL = 7D H

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