Question

Using emu8086.inc , write an assembly program that display the message

'Enter the number:'

Then, reads a byte from the user, then count the number of 1’s in it, and display them on screen.

Answer 1

Hello!

EXPLAINATION:

Before going to the code first we will look into some basics of 8086

AX, BX , CX and DX are 16 bit general purpose registers.

AX is also called as accumulator. any operation like division and ultiplication the result is stored in accumulator(AX)

We can have have 8 bits by refering X with L or H like AH and AL both are 8 bit registers AX = AH + AL , where H represent Higher byte L represents lower byte.

MOV operand 1, operand 2

In this instruction first operand mush be register(AX,AL,AH,BX...) or memory location([DI],[SI]) , second operand can be immediate data or register or memory location.

Ex:

MOV AX, 0015h

MOV [4000h],12h

DB , DW are directives

Directives are used to store inforamtion in the memory

Here i have taken org 100h the program is utilising after first 2 bytes so now our 1st word starts storing at 102h offset address.

var_name db value1,value2,.......value n

Byte(8bits)

In 8086 you can store the data into memory using variables here var_name is a variable so it will allocate the 1 byte memory to the values

var_name dw value1,value2,.......value n

Word(16bits or 2 bytes)

Here var_name is a variable so it will allocate the 2 bytes memory to the all the values

var_name db ?

This creates a variable with no data assigned to it so use ? when dont have clue store which value in varaible.

NOTE: values can be number or charecters or string

VAR_NAME , var_name both are same, 8086 is not case sensitive

ADD operand 1, opearand 2

This instruction add the operand 1 and operand 2 and result stores back to the operand 1.

In this instruction operand 1 must not be immediate value but it can be a register and memory location.

Operand 2 can be any thing register or memory location or register.

Ex:

ADD Cl , 05h

ADD [SI] , 45h

ADD 78h, AX ; first operand must not be immediate value

ADD [DI],Var1 ; both operands must not be addresses

Procedures:

Procedures helps us in code reusability , its similar to functions sometimes same instructions we need to execute for different data in that situation we can use procedures

SYNTAX :

PROC procedure_name

; code

RET

ENDP

INTERUPTS:

Interrupts will stop the main program and executes a sub routine(other than main program) and comes back to the main progarm and continues its execution after the interrupt.

There are so many sub routines so in order to execute the particular sub routine so value we should give to the AX register before interrupt.  

There are lot of intrupts in 8086 , and to discuss here about all is out of the sope of this question.

If you want to know more about interrupts please go to the HELP section in EMU8086 then go to documentation index page there you can find interrupts

Loding offset address:

When enter is a variable declared above in DATA segments then we can get offset address in 2 ways

LEA DX,enter

MOV DX,offset enter

both are same.

STACK operations:

Stack is contiguous memory allocation we can push or pop only top of the stack .

PUSH BX means we are pushing the value in BX to stack

POP BX means we are getting the data from top stack of stack and stores in the register BX , by doing this in stack the data will be deleted and comes to register.

Algorithm used:

First i have stored the strings like 'Enter the number ', 'Number of ones are:-' and new line charecter to respentive varibles.

Count variable is used to store the number 1s in the given byte

First it will display the msg Enter the number and then it takes the bit untill 8 bits are entered.

Each time when we enter bit it will be stored as ascii format suppose we enter 1 means then 31 is pushed to stack, so we will subtract 30 and then push to the stack.

Next it will display msg number of ones are:-

And then it will count number of 1s by poping from the stack , at this part we will check whether it is 1, if it is 1 then we will increment DH , so finally DH stores the number of 1s in the byte.

Finally Displaying number of 1s occured to the screen.

CODE:

.DATA

   enter DB 0DH,0AH, "ENTER THE NUMBER :-$"
   number DB 0DH,0AH, "NUMBER of one's are :- $"
   new_line db 0dh,0ah,"$"  
   count DB ?

.CODE
   START:

   MOV AX,@DATA
   MOV DS,AX
              
   ; Displays 'Enter the number'            
              
   ENTER_MSG:
   LEA DX,enter     ; Offset address of enter variable
   MOV AH,09H       ; display subroutine
   INT 21H          ; call the subroutine by intrupting
   MOV CL,00        ; make value in cx is 0
   MOV AH,01H       ; set the cursor at 1st line
   
   
   CALL INPUT_BYTE  ; call the input string procedure
   
   ; Displays 'Number of one's are:-'
   
   COUNT_MSG:       ; lable for displaying reverse msg
   
   LEA DX,number    ; Dx gets stored with offset address of reverse variable
   MOV AH,09H       ; Display subroutine
   INT 21H          ; call the subroutine by intrupting
   lea dx, new_line ; DX gets stored with offset address of new_line variable
   mov ah,09h       ; Display subroutine
   int 21h          ; call the subroutine by intrupting
   
   MOV DH,00h
   
   ; Counts number of 1s in given byte
   
   COUNT_ONEs:
   
   POP BX           ; pops the last entered value and stores back to BX
   MOV DL,BL        ; Move BL value to DL
   POP BX                                    
   CMP DL,01h       ; Compare the bit wether it is 1 or not 
   JNZ skip         ; If 1 then execute the below instruction
   INC DH           ; increment the DH value   
   skip:
   LOOP COUNT_ONEs  ; loop till the value in CX is 0  
   
   ; Display final count of 1s to the screen
   
   mov al,dh        ; convert to binary coded decimal,
   
   or al, 30h       ; convert to printable symbol:

   ; print char in al using bios teletype function:
   mov ah,0eh
   int 10h


 

HLT                 ; HALT The program

INPUT_BYTE PROC     ; procedure to take input from keyboard

    INT 21H         ; call the subroutine by intrupting
    MOV BL,AL       ; moves value in AL to BL   
    SUB BX,30h      ; Subtract 30 as ascii value for numbers starts from 30-39
    PUSH BX         ; push the BX to stack
    INC CL          ; calculate number of bits
    CMP CL,08h      ; If user enters 8 bits
    JZ COUNT_MSG    ; then goes to count_msg
    CALL INPUT_BYTE ; otherwise it calls the same procedure till user enters 8 bits    
    RET             ; RETURN back to the caller

INPUT_BYTE ENDP     ; End of procedure

.EXIT

   END  START

edit: C:\emu8086\MySource\number_of_ones_in.asm file edit bookmarks assembler emulator math ascii codes help ? help E about new open examples save compile emulate calculator convertor options 01 .DATA 02 03 enter DB ODH,GAH, "ENTER THE NUMBER :-$" 94 number DB ODH, DAH, "NUMBER of one's are : - $" 05 new_line db Odh, ah, "$" 06 count DB ? 07 08 .CODE 99 START: 19 11 MOU AX, DATA 12 MOU DS, AX 13 14 ; Displays 'Enter the number' 15 16 ENTER_MSG: 17 LEA DA, enter ; Offset address of enter variable 18 MOV AH,09H ; display subroutine 19 INT 21H call the subroutine by intrupting 20 MOU CL,00 make value in cx is 0 21 MOV AH,01H set the cursor at 1st line CALL INPUT_BYTE ; call the input string procedure ; Displays Number of one's are:- COUNT_MSG: ; lable for displaying reverse msg LEA DX, number Dx gets stored with offset address of reverse variable MOV AH,09H ; Display subroutine INT 21H ; call the subroutine by intrupting lea dx, new_line DX gets stored with offset address of new_line variable mov ah,09h ; Display subroutine int 21 h call the subroutine by intrupting MOU DH,00 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ; Counts number of 1s in given byte COUNT_ONES: POP BX ; pops the last entered value and stores back to BX MOV DL, BL ; Move BL value to DL POP BX CMP DL,91 h ; Compare the bit wether it is 1 or not JNZ skip ; If i then execute the below instruction INC DH ; increment the DH value skip: LOOP COUNT_ONES ; loop till the value in CX is o ; Display final count of 1s to the screen moval,dh ; convert to binary coded decimal, or al, 30h ; convert to printable synbol: ; print char in al using bios teletype function: 54 55 56 57. 58 50 รห oh

edit: C:\emu8086\MySource\number_of_ones_in.asm file edit bookmarks assembler emulator math ascii codes help ? E new open examples save compile emulate calculator convertor options help about 49 skip : 50 LOOP COUNT _ONES ; loop till the value in CX is 0 51 52 ; Display final count of 1s to the screen 54 mov al,dh convert to binary coded decimal, 55 56 or al, 30h ; convert to printable synbo 1: 57 58 ; print char in al using bios teletype function: 59 mov ah, eh 60 int 10h 61 62 63 64 65 HLT ; HALT The program 66 67 INPUT_BYTE PROC ; procedure to take input from keyboard 68 69 INT 21H ; call the subroutine by intrupting 70 MOU BL, AL moves value in AL to BL 71 SUB BX,30h Subtract 30 as ascii value for numbers starts from 30-39 72 PUSH BX push the BX to stack 73 INC CL calculate number of bits 74 CMP CL,08h If user enters 8 bits 75 JZ COUNT_MSG then goes to count_msg 76 CALL INPUT_BYTE otherwise it calls the same procedure till user enters 8 bits 77 RET ; RETURN back to the caller 78 79 INPUT_BYTE ENDP ; End of procedure 80 .EXIT 82 83 END START 84

OUTPUT:

X 07 08 .CODE 09 10 SCR emulator screen (80x25 chars) 11 12 13 ENTER THE NUMBER:-00001111 14 NUMBER of one's are:- 15 4 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 clear screen change font ab 0/16

-CODE SCR emulator screen (80x25 chars) ENTER THE NUMBER:-11111111 NUMBER of one's are:- 8 1/16 clear screen change font 49

CODE SC emulator screen (80x25 chars) х ENTER THE NUMBER:-00000000 NUMBER of one's are:- 0 0/16 clear screen change font VU UTL TT UTTUGI I ITT SIVGI WYLE

Х 98 .CODE 19 SCR emulator screen (80x25 chars) 1 2 3 ENTER THE NUMBER:-00011100 -4 NUMBER of one's are:- 5 3 .6 7 8 9 21 2 23 24 25 6 7 28 29 1 2 3 34 5 36 7 38 1/16 clear screen change font 48

AFTER EXECUTION:

E about IN IR SCR emulator screen (80x25 chars) х ENTER THE NUMBER:-01110001 NUMBER of one's 4 10 are: om 39-1 edit: C:\emu8086\MySource\number_of_ones_in.asm file edit bookmarks assembler emulator math ascii codes help original source co... ? new open examples save compile 51 elp 49 skip : 52 ; Display final count of 50 LOOP COUNT _ONES ; loop till 53 51 54 mov al,dh conve 52 ; Display final count of 1s 155 56 oral, 30h conve 54 mov al,dh convert td 57 55 58 ; print char in al using 56 or al, 30h ; convert td 59 mov ah, eh 57 60 int 10h 58 ; print char in al using bios 61 59 mou ah, eh 62 60 int 10h 63 61 64 62 65 HLT ; HA , 63 64 65 HLT ; HALT The program 66 67 INPUT_BYTE PROC mrocedure to take innut from ke uhoard 68 69 INT emulator number_of_ones_in.exe_ 70 MOV 71 SUB file math debug view external virtual devices virtual drive help 72 PUS 73 INC I ID 74 CMP Load reload step back single step run 75 JZ step delay ms: 0 76 CAL RET registers H L 0714:9038 9714: 0038 78 79 INPUT_B AX SE 34 07178: F4 244 HLT 80 071798 CD 205 = INT 021h .EXIT B8 | 00 13 0712A: 21 033 ! MOU BL, AL 82 0717B: 8A 138 è SUB BX, 030h 83 END OG OO 0712C: D8 216 + PUSH BX 84 0717D: 83 131 a INC CL DX 9400 0717E: EB 235 6 CMP CL, 08h 0717F: 30 048 JZ 0113h CS 6714 07180: 53 983 S CALL 00039h 07181: FE 254 IP RET 0938 07182: C1 193 MOU AX, 94000h SS 07183: 80 128 Ç INT 921h 6716 07184: F9 249 SP OSOS 07185: 98 008 BACK NOP 07186: 74 116 t NOP BP |ទ្ធិទទ្ធ 07187: CB 203 NOP 07188: E8 232 0 NOP SI 9996 07189: EE 238 € NOP 0718A: FF 255 RES NOP DI GGGG 0718B: C3 195 F NOP 0718C: B8 184 1 NOP DS 0716 0718D: 00 900 NULL ES 0706 screen source reset aux vars debug stack flags enters 0/16 clear screen NOP change font D

Hope this helps and clear!

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Thank you!