Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

check for these assembly code, why cant it compile in dosbox \"tasm\". what is m

ID: 3767296 • Letter: C

Question

check for these assembly code, why cant it compile in dosbox "tasm". what is missing?

.Model small
.stack 100h
.code

BASE64 DB "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

DB "abcdefghijklmnopqrstuvwxyz0123456789+/"

BUFFER DB '000','$'

mov ecx, L

mov esi, sourc

;data from source table

mov edi, character

push ebp

mov ebp, Bufferout

source:

xor eax, eax

;Reading data using Source table

mov ah, byte ptr[esi]

mov al, byte ptr[esi+1]

;readingnext address

shl eax, 16

mov ah, byte ptr[esi+2]

;edx bit is set to 1

mov edx, eax

shl eax, 6   

;first 6 bits

shr edx, 26   

mov bl, byte ptr [edi+edx]

;put data or character in buffer

mov byte ptr[ebp], bl

inc ebp   

;next value from the buffer

mov edx, eax

shl eax, 6   

;next 6 bits

shr edx, 26

mov bl, byte ptr [edi+edx]   

;Takes first character from the buffer

mov byte ptr[ebp], bl

inc ebp

;takes next character from the buffer

mov edx, eax

shl eax, 6   

;done first 6 bits

shr edx, 26

mov bl, byte ptr [edi+edx]

;put char in buffer

mov byte ptr[ebp], bl

inc ebp

;manipulate in edx bitset4

mov edx, eax

shl eax, 6   

;done first 6 bits

shr edx, 26

mov bl, byte ptr [edi+edx]   

;Character is stored in the buffer

mov byte ptr[ebp], bl

inc ebp   

;next value stored in the buffer

add esi, 3

sub ecx, 3

cmp ecx, 3

jge source   

;information still present to convert

xor eax, eax   

;padding to zero

cmp ecx, 0

jz finish

;reading next data

mov ah, byte ptr[esi]

mov al, byte ptr[esi+1]

shl eax, 16

mov ah, byte ptr[esi+2]

;information read

sub ecx, 3   

;+ve inverse

neg ecx   

;byte information which needs padding

mov edx, ecx   

;padded bits equated to zero

mov esi, 0xFFFFFF

;calculating number of shifts to convert to base64

lea ecx, dword ptr[ecx*8+8]

shl esi, cl

;masking unwanted data

and eax, esi

mov ecx, edx   

;byte 1 in edx

mov edx, eax

shl eax, 6

shr edx, 26

;storing next data into the buffer

mov bl, byte ptr [edi+edx]   

mov byte ptr[ebp], bl

;next bufer value in ebp

inc ebp   

;byte 2 in edx

mov edx, eax

shl eax, 6

shr edx, 26

  

;Stores characters into the buffer

mov bl, byte ptr [edi+edx]   

mov byte ptr[ebp], bl

inc ebp

;byte 3 in edx

mov edx, eax

shl eax, 6

shr edx, 26

;stores character into the buffer

mov bl, byte ptr [edi+edx]   

mov byte ptr[ebp], bl

;Next buffer

inc ebp   

;byte 3 in edx

mov edx, eax

shl eax, 6

shr edx, 26

;Storing char into buffer

mov bl, byte ptr [edi+edx]   

mov byte ptr[ebp], bl

;next avlue in the buffer

inc ebp   

;'returns' padded count

mov eax, ecx

finish:

test eax, eax

jz end

  

;padded data =

sub ebp, eax   

;CharacterPadded:

mov byte ptr[ebp], 0x3d

inc ebp

dec eax

jnz CharacterPadded

end:
  

Explanation / Answer

Erroe in the given program :

missing DB represents Define byte which is used to declare a byte type variable or to store a byte in memory location.

For example

Byte 1 DB 10h

Byte 2 DB 255 which is equal to 0ffh and this is the maximum possible value for a given byte

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Chat Now And Get Quote