Can you also explain what each line of code does in order to reach the displayed

Question

Can you also explain what each line of code does in order to reach the displayed ASCII character?
thanks!

What will be displayed on the screen? Explain in detailst 32 20 040 Sp 33 21 041 ! 34 22 042 34; " 35 23 043 "#35; # 36 24 044 $ 37 25 045 6#37 ; 38 26 046 & 39 27 047 ' 40 28 050 ( ( 41 29 051 ) ) 42 2A 052 42; 43 2B 053 + 44 2C 054 , 45 2D 055 G#45; 46 2E 056 . 47 2F 057 6#47: 48 30 060 0 .model small . stack 00h 1 . code main proc mov dl, 191 shr dl, 2 f: inc dl inc dl call t cmp dl, 65h 49 31 061 G#49; 50 32 062 2 51 33 063 3 52 34 064 52 mov ax, 4c00h int 21h t proc 53 35 065 5 54 36 066 6 mov ah, 6 int 21h ret 55 37 067 7 56 38 070 56 57 39 071 57 58 3A 072 : 59 3B 073 7 60 3C 074 = 61 3D 075  62 3E 076  main endp end main

Explanation / Answer

Code: (Explained line by line)

.model small
.stack 100h

.code

main proc

mov dl, 191 ;moves 191 into lower 8 bits of DX register
shr dl, 1 ;logical right shift dl by 1, now dl = 95

f: ; loop f

inc dl ; increment dl, dl = 96
inc dl ; increment dl, dl = 97
call t ; call loop t
cmp dl, 65h ; compare dl to 65h or 101
js f ; jumps back to f if sign flag is set
  
mov ax, 4c00h ; moves 0x4c00 into AX register
int 21h ; because AH = 0x4C, program exits
t proc
mov ah, 6 ; print value in dl to output
int 21h ; call 21h interrupt
ret ; return to caller function
t endp

main endp
end main

Till dl register is less than 101, the program will stay in loop f because dl - 65h is negative and the sign flag is set till dl becomes 101 or 65h.

The output will be ace

Hope this helps. Do Upvote! :)

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