如何换行并在换行中打印相同的字符

Question

I'm trying to learn assembly. I saw this example in printing "Hello World!(red text) with a backgound color(yellow)"

I managed to edit the code to just print spaces with yellow background by trial and error. However I cannot print a new line. if I add a new mov [200], ' ' for example (dont know if this is correct) it adds a character on a different line but a different color.. if I add 00010001b after the comma if prints a different color which should blue.

can anyone give me a head start tutorial in this code. I just want to print another line for now.. Here is the working code so far.. it prints a whole line of yellow

name "hi-world"


; hex    bin        color ;  ; 0      0000      black ; 1      0001    blue ; 2      0010      green ; 3      0011      cyan ; 4      0100    red ; 5      0101      magenta ; 6      0110      brown ; 7      0111  light gray ; 8      1000      dark gray ; 9      1001      light blue ; a      1010      light green ; b      1011      light cyan ; c      1100      light red ; d      1101      light magenta ; e      1110     yellow ; f      1111      white



org 100h

; set video mode     mov ax, 3     ; text mode 80x25, 16 colors, 8 pages (ah=0, al=3) int 10h       ; do it!

; cancel blinking and enable all 16 colors: mov ax, 1003h mov bx, 0 int 10h


; set segment register: mov     ax, 0b800h mov     ds, ax

; print "hello world" ; first byte is ascii code, second byte is color code.

mov [02h], ' '

mov [04h], ' '

mov [06h], ' '

mov [08h], ' '

mov [0ah], ' '

mov [0ch], ' '

mov [0eh], ' '   

mov [10h], ' '

mov [12h], ' '

mov [14h], ' '

mov [16h], ' '

mov [18h], ' '

mov [1ah], ' '

mov [1ch], ' '

mov [1eh], ' '   

mov [20h], ' '


; color all characters: mov cx, 34  ; number of characters. mov di, 03h ; start from byte after 'h'

c:  mov [di], 11101100b   ; light red(1100) on yellow(1110)
    add di, 2 ; skip over next ascii code in vga memory.
    loop c

; wait for any key press: mov ah, 0 int 16h

ret

Answer 1

When you write directly to video memory at B800:<adr> , the address of write decides what the value will be used for (depends on selected graphics(text!) mode of course).

In classic text mode 80x25 the video memory starts at B800:0000 , and has size 80*25*2 bytes. 80*25 is probably self explanatory, 1 byte per character (in extended 8-bit ASCII encoding), so why *2? Each character has 1 more byte dedicated for colours. Actually they sit next to each other, so at B800:0000 is the ASCII value of character, and at B800:0001 is stored it's colour attribute.

Character+attribute pairs are stored from left to right row by row, so to write character '*' at position (40,12) (almost centre of screen), you have to write at address (y*160 + x*2) = (12*160 + 40*2) = 2000. The *160 is *80*2 = number of character+attribute pairs per row, ie. size of row in bytes:

mov   BYTE PTR [2000],'*'
mov   BYTE PTR [2001],2Eh  ; yellow ink, green paper

Or you can shorten that to single WORD write as:

mov   WORD PTR [2000],2E00h + '*'

To print on next line you simply have to adjust the address either by +160 (to move under current character), or by adding the remaining size of row to the current address to get to the first char of next line, like +80 for the 2000 from example to move at the beginning of line 13 (14th line on screen).

See this page for further details about video memory layout:
http://www.shikadi.net/moddingwiki/B800_Text

---

BTW the more bytes you write at one time, the faster the code runs on real (original) HW, so in your case I would strongly prefer either the WORD write storing the ASCII+attribute pair together, or even DWORD writes (in 32b mode) storing two letters+two attributes with single instruction.

Your code does first write every second byte on even addresses setting the ASCII letters, then it does write every odd byte with colour value. I would do it in this way personally:

mov ax,0B800h
mov es,ax
mov di,<target_address = y*160 + x*2>  ; es:di set up
mov ah,<color_attribute_for_whole_string>
; now the print by letter gets here
mov al,'H'  ; al = letter, ah = attribute (ax set up)
stosw       ; [es:di] = ax, di += 2
mov al,'e'  ; change only letter in ax
stosw
mov al,'l'
stosw
mov al,'l'
stosw
...

; the mov al,<char> can be replaced by LODSB
; reading ascii string from ds:si.
; then it would look like:
mov   si,<address_of_ascii_string>
lodsb
stosw
lodsb
stosw
... so many times as the string length