EMU8086 Display screen (black screen)

Question

I am taking Microprocessors course and I have a problem with my code.

I need to write a program code that make a word moving around.

The code work correctly on (80*25) screen (because of my calculations depends on it)

But the problem is: when I maximize the display screen everything goes wrong!, because the dimensions changed

Is there any instruction code that if we write in the code enforce the display to contain the same dimensions even if we maximize it?

when the display screen (80*25)

screen Maximized (191*63)!

    mov ax, 0b800h

    mov ds, ax
    mov bx, 0 
    Main:    
    call dltru


    dltru:   ; diagonal left to right upper   
    mov bx,0
    mov si,0
    mov cx,21
    lower2:  
    call box         
    mov [bx+160+2+si],' '       
    mov [bx+160+4+si],'M'
    mov [bx+160+6+si],'O'
    mov [bx+160+8+si],'E'
    mov [bx+160+10+si],' '
    mov [bx+160+12+si],'S' 
    mov [bx+160+14+si],'A'       
    mov [bx+160+16+si],'Y'
    mov [bx+160+18+si],'E'
    mov [bx+160+20+si],'L'
    mov [bx+160+22+si],' '
    mov [bx+160+24+si],'F' 
    mov [bx+160+26+si],'A'
    mov [bx+160+28+si],'L'
    mov [bx+160+30+si],'A'
    mov [bx+160+32+si],'H'
    mov [bx+160+34+si],' '           
    call delete            
    add si,160
    add bx,6
    loop lower2    
    ret    


    Box: ;this will draw the box around the announcement!        
    ;left
    mov [0+bx+si],0B2h
    mov [160+bx+si],0B2h
    mov [320+bx+si],0B2h         
    ;right
    mov [36+bx+si],0B2h
    mov [196+bx+si],0B2h
    mov [356+bx+si],0B2h
;upper
mov [2+bx+si],0B2h
mov [4+bx+si],0B2h
mov [6+bx+si],0B2h 
mov [8+bx+si],0B2h
mov [10+bx+si],0B2h
mov [12+bx+si],0B2h  
mov [14+bx+si],0B2h
mov [16+bx+si],0B2h
mov [18+bx+si],0B2h 
mov [20+bx+si],0B2h
mov [22+bx+si],0B2h
mov [24+bx+si],0B2h 
mov [26+bx+si],0B2h
mov [28+bx+si],0B2h
mov [30+bx+si],0B2h
mov [32+bx+si],0B2h
mov [34+bx+si],0B2h       
;lower
mov [322+bx+si],0B2h
mov [324+bx+si],0B2h
mov [326+bx+si],0B2h 
mov [328+bx+si],0B2h
mov [330+bx+si],0B2h
mov [332+bx+si],0B2h  
mov [334+bx+si],0B2h
mov [336+bx+si],0B2h
mov [338+bx+si],0B2h  
mov [340+bx+si],0B2h 
mov [342+bx+si],0B2h
mov [344+bx+si],0B2h
mov [346+bx+si],0B2h
mov [348+bx+si],0B2h
mov [350+bx+si],0B2h
mov [352+bx+si],0B2h
mov [354+bx+si],0B2h       
ret


delete:  
mov [bx+162+si],' '
mov [bx+164+si],' '
mov [bx+166+si],' '
mov [bx+168+si],' '
mov [bx+170+si],' '
mov [bx+172+si],' '
mov [bx+174+si],' '
mov [bx+176+si],' '
mov [bx+178+si],' '
mov [bx+180+si],' '
mov [bx+182+si],' '
mov [bx+184+si],' '
mov [bx+186+si],' '
mov [bx+188+si],' '     
mov [bx+190+si],' '
mov [bx+192+si],' '
mov [bx+194+si],' '     
;left
mov [0+bx+si],' '
mov [160+bx+si],' '
mov [320+bx+si],' '         
;right
mov [36+bx+si],' '
mov [196+bx+si],' '
mov [356+bx+si],' '
;upper
mov [2+bx+si],' '
mov [4+bx+si],' '
mov [6+bx+si],' ' 
mov [8+bx+si],' '
mov [10+bx+si],' '
mov [12+bx+si],' '  
mov [14+bx+si],' '
mov [16+bx+si],' '
mov [18+bx+si],' ' 
mov [20+bx+si],' '
mov [22+bx+si],' '
mov [24+bx+si],' ' 
mov [26+bx+si],' '
mov [28+bx+si],' '
mov [30+bx+si],' '
mov [32+bx+si],' '
mov [34+bx+si],' '       
;lower
mov [322+bx+si],' '
mov [324+bx+si],' '
mov [326+bx+si],' ' 
mov [328+bx+si],' '
mov [330+bx+si],' '
mov [332+bx+si],' '  
mov [334+bx+si],' '
mov [336+bx+si],' '
mov [338+bx+si],' '  
mov [340+bx+si],' ' 
mov [342+bx+si],' '
mov [344+bx+si],' '
mov [346+bx+si],' '
mov [348+bx+si],' '
mov [350+bx+si],' '
mov [352+bx+si],' '
mov [354+bx+si],' '       
ret

Answer 1

mov [bx+160+2+si],' '

Because of the fixed value of 160, that is only valid for an 80 columns screen, your code fails on the alternative resolution.

What you need is calculating the size of 1 scanline in the textvideo memory:

. First look up the actual number of columns on the current screen. BIOS has this count in the memory variable at 0040h:004Ah.
. Then double this value because in the video memory each character/attribute occupies 2 bytes.
. Finally use this value instead of the fixed value of 160.

push ds
mov  bx, 0040h
mov  ds, bx
mov  bx, [004Ah]
shl  bx, 1     ; This gives: BX=160 if 80x25, BX=382 if 191x63
pop  ds

Writing on the 2nd row of the screen becomes:

mov  [bx+2+si],' '       
mov  [bx+4+si],'M'
mov  [bx+6+si],'O'
mov  [bx+8+si],'E'

Writing on the 3rd row of the screen becomes:

add  bx, bx
mov  [bx+2+si],0B2h
mov  [bx+4+si],0B2h
mov  [bx+6+si],0B2h 
mov  [bx+8+si],0B2h