Question

Hi this is Assembly Language MASM x86 program. Please write it in the language and please explain it with comments thank you

Please answer it I really need help this question was refunded before so please answer. Thank you so much also these are two separate programs thank you.

1) Write a procedure to read in decimal or hex number (byte-sized) Then write a procedure using shifts and ANDS to convert the string to a binary number (if is backward, that ok)

2) Put the first (byte) 10 terms of the Fibonacci series into the byte array FIB 1 1 2 3 5 … Us num1 = 1 and num2 = 1 to start, NO other numbers allowed)

Answer 1

1) The code with detailed comments is given below:

assume cs:code, ds:data

;define data segment
data segment          
    msg1 db "Enter the Hexadecimal number:$" ; message to read a hexadecimal number
    str db 3,0,3 dup(?) ;memory location storing details on the number read
    ; byte sized hexadecimal number can hold 2 hex characters. Hence max size of str is 3 where it can hold
    ; 2 characters with an additional space for enter key.
    msg2 db 10,13,"The binary representation is: $" ; message to display binary representation of the number
data ends

code segment
start:
    mov ax, data
    mov ds, ax ; initialization of data segment
          
    lea dx, msg1 ; display msg1
    mov ah, 09h
    int 21h
  
    lea dx,str ; read the number using buffered input
    mov ah, 0Ah
    int 21h
  
    lea dx, msg2 ; display msg2
    mov ah, 09h
    int 21h
  
    call HexToBin
         
    mov ah, 4ch ; terminate the program
    int 21h  

HexToBin PROC
    mov cl,str+1 ; in case of buffered input, the length of input will be stored at location str+1
    ; cl register holds the length of the string read
    mov ch,00 ; ch stores the value of 0
    lea si,str+2 ; the strating address of the number read is stored in SI index register
    ; The following steps are repeated for every character in the number read
up:
    mov al,[si]; character is read from location memory location pointed to by SI
    ; The ASCII values of Hex numbers can range from 30H to 39H and 41H to 46H
    cmp al,39h; check if ascii value is below or equal to 39h
    ; if the ascii code is <=39H then just and it with 0FH to obtain numbers from 01 to 09H respectively
    jbe next
    ; if ascii value id greater than 39H, then is ranging from A to F
    sub al,07h ; first subtact 7 from it to obtain values in range 3AH to 3FH
    ; then AND it with 0FH to obtain 0AH to 0FH
next:
    and al,0Fh
    ;we would like to have only the 4 digit binary representation for each digit. So the higher order 4 bits is
    ; not necessary
    shl al,4 ; first we shift contents of al left by 4 bit positions to remove the higher order 4 bits     
    mov bl,04; the remaining 4 bits is printed one by one in a loop and bl is loop counter
print:
    mov dl,30h ; dl stores the ascii code of 0
    shl al,1 ; shift left is done to print the binary representation if the proper order.
    ; If reverse representation is required, shift right can be used.
    adc dl,0; the shifted bit will be in the CF; if there is a carry, adc will add 1 to dl else dl will remain 30H
    ; Hence, if there is carry, 1 will be printed on screen and if there is no carry, 0 will be printed on screen
    push ax ; push ax to prevent change in value of al
    mov ah,02h ; print the value in dl on screen
    int 21h
    pop ax ; pop ax will restore value of al
    dec bl ; decrement loop counter
    jnz print; repeat loop until bl is zero
    inc si ; to print next character, increment si pointer    
    loop up ; outer loop continues until cx!=0
    ret
HexToBin ENDP

code ends   
end start

OUTPUT:

2) The program for computing 10 Fibonacci numbers is as follows:

data segment      
    num1 db 1
    num2 db 1
    fib db 10 dup(?)
data ends
code segment
start:
    mov ax, data
    mov ds, ax ; initialize the data segment  
    lea si,fib ; make SI pointer point to the fib byte array
    mov al,num1 ; store num1 as 1st element in fib array
    mov [si],al
    inc si ; increment si to point to next memory location
    ;store num2 as 2nd element in fib array
    mov al,num2
    mov [si],al
    inc si ; incement si pointer
    mov cx,8 ; since two numbers are already stored, 8 more numbers will have to be generated.
up:
    mov al,num1 ; mov num1 to al
    add al,num2 ; add num2 to al
    daa ; convert the result of hexadecimal addition to decimal value by using DAA
    mov [si],al ; store the result in location pointed to bi SI
    ; perform, num1 = num2, num2 = al using register bl
    mov bl,num2
    mov num1,bl
    mov num2,al
    inc si ; increment si pointer
    loop up ; execute the loop until 8 fibonacci numbers are generated.
    mov ah, 4ch ; exit to operating system.
    int 21h  
code ends
end start

OUTPUT: