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Assignment No.2                                                                                    Dated:Jan 12, 18

Dear Students,

Assignment No.2 has been uploaded; its due date is Thursday, 18- January -2018.

Note 1: Assignment will not be accepted via email in any case. Upload your assignment on VU-LMS with in due date.

Note 2: All times on VU-LMS as according to Pakistan Standard Time (PKT)

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yei problem aa raha

CS401 Assignment no.2 Solution step to step

https://youtu.be/aLoXQAIGvz0

Assignment solution jo me ny bnaya he wo kuch is trha he.

step 1. Write clear screen routine given in handsout (example 6.2)

step 2. Write kbisr routine code from handsout (example 9.6)

Is example me jaha left aur right keys press hen waha apni id ka code fit krny ki koshish kro.

step 3. Then next step is to move screen position 1 step forward with every key stroke, and that is the most challenging part of assignment.  

Yaar ap apna code muje is link me pvt send ker do me ap ki id wagera ko copy nahi karo ga bs code deekho ga or mere wale mein jo kami ha woo sahi karo ga ... ap ka code leek nahi ho ga is tara ... https://www.facebook.com/halibhutta

Array bhai code leak ki bat nhi he me help krna chahta hoon lkn yaha to logon ki bohat hi buri halat he.

me apna code yaha hi share kr deta hoon apni id hata ke

shoaib  bhai  tareka send kar dy 

wrong hai yeh bhi

nasm ka use he bta dy

; clear screen using string instructions; clear screen using string instructions

[org 0x0100] jmp start oldisr: dd 0 ; space for saving old isrnext: dw 0 ; subroutine to clear the screenclrscr: push es push ax push cx push di mov ax, 0xb800 mov es, ax ; point es to video base xor di, di ; point di to top left column mov ax, 0x0720 ; space char in normal attribute mov cx, 2000 ; number of screen locations cld ; auto increment mode rep stosw ; clear the whole screen pop di pop cx pop ax pop es ret
; keyboard interrupt service routine
kbisr: push ax push es

mov ax, 0xb800 mov es, ax ; point es to video memory in al, 0x60 ; read a char from keyboard port nextcmp1:cmp al, ---- ; has the - key pressed jne nextcmp2 ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nextcmp2: cmp al, ---- ; has the - key pressed jne nextcmp3 ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nextcmp3: cmp al, ---- ; has the - pressed jne nextcmp4 ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nextcmp4: cmp al, ---- ; has the - pressed jne nextcmp5 ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nextcmp5: cmp al, ---- ; has the - pressed jne nextcmp6 ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nextcmp6: cmp al, ---- ; has the 2 pressed jne nextcmp7 ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nextcmp7: cmp al, ---- ; has the 4 pressed jne nomatch ; no, try next comparison mov word [es:di+2], ---- ; yes, print word on screen add di, 2 jmp exit ; leave interrupt routine

nomatch: pop es pop ax jmp far [cs:oldisr] ; call the original ISR
exit: mov al, 0x20 out 0x20, al ; send EOI to PIC pop es pop ax iret ; return from interrupt
start: call clrscr ; call clrscr subroutine xor ax, ax mov es, ax ; point es to IVT base mov ax, [es:9*4] mov [oldisr], ax ; save offset of old routine mov ax, [es:9*4+2] mov [oldisr+2], ax ; save segment of old routine cli ; disable interrupts mov word [es:9*4], kbisr ; store offset at n*4 mov [es:9*4+2], cs ; store segment at n*4+2 sti ; enable interrupts
l1: mov ah, 0 ; service 0 – get keystroke int 0x16 ; call BIOS keyboard service cmp al, 13 ; is the Ent key pressed jne l1 ; if no, check for next key mov ax, [oldisr] ; read old offset in ax mov bx, [oldisr+2] ; read old segment in bx cli ; disable interrupts mov [es:9*4], ax ; restore old offset from ax mov [es:9*4+2], bx ; restore old segment from bx sti mov ax, 0x4c00 ; terminate program int 0x21

Lo g bhaio is code me jaha jaha ---- lga hoa he waha apni id fit kr lo. mery pas ye code sahi kam kr rha he, aagy ab hr ksi ki apni apni himat he, koi guidance chaie ho to contact me. 

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