Jan db " January$ "
string db "Sun Mon Tue Wed Thu Fri Sat$"
string1 db " 1 2 3$"
string2 db " 4 5 6 7 8 9 10$"
string3 db "11 12 13 14 15 16 17$"
string4 db "18 19 20 21 22 23 24$"
string5 db "25 26 27 28 29 30 31$"
There are several ways to achieve your goal IN TEXT MODE :
- Display char by char : this way you can choose one color for every character.
- Access screen memory : at segment 0B800:0.
- Display the whole string with the same color.
Next code does the job with the third option (the easiest). It was made with EMU8086:
.stack 100h
.data
Jan db " January ",13,10
db "Sun Mon Tue Wed Thu Fri Sat",13,10
db " 1 2 3 ",13,10
db " 4 5 6 7 8 9 10 ",13,10
db "11 12 13 14 15 16 17 ",13,10
db "18 19 20 21 22 23 24 ",13,10
db "25 26 27 28 29 30 31 "
color db 181
.code
;INITIALIZE DATA SEGMENT.
mov ax,@data
mov ds,ax
;DISPLAY STRING WITH COLOR.
mov es,ax ;ES SEGMENT MUST POINT TO DATA SEGMENT.
mov ah,13h ;SERVICE TO DISPLAY STRING WITH COLOR.
mov bp,offset Jan ;STRING TO DISPLAY.
mov bh,0 ;PAGE (ALWAYS ZERO).
mov bl,color
mov cx,201 ;STRING LENGTH.
mov dl,0 ;X (SCREEN COORDINATE).
mov dh,5 ;Y (SCREEN COORDINATE).
int 10h ;BIOS SCREEN SERVICES.
;FINISH THE PROGRAM PROPERLY.
mov ax,4c00h
int 21h
Notice I removed the $ signs (because 13h service requires string length, not $). For a different color just change the value (181) for the "color" variable in data segment.
To display diferent colors for diferent strings, copy-paste the display block for every string.
Let us know if it worked for you.
The formula to choose color goes like this:
text-background * 16 + text-color
Next are the colors :
Black = 0
Blue = 1
Green = 2
Cyan = 3
Red = 4
Magenta = 5
Brown = 6
LightGray = 7
DarkGray = 8
LightBlue = 9
LightGreen = 10
LightCyan = 11
LightRed = 12
LightMagenta = 13
Yellow = 14
White = 15
With the given formula, if you want red background with yellow text you need the color 78 :
4 * 16 + 14 = 78
Now, let's do it in GRAPHICS MODE :
.stack 100h
.data
Jan db " January ",13
db "Sun Mon Tue Wed Thu Fri Sat",13
db " 1 2 3 ",13
db " 4 5 6 7 8 9 10 ",13
db "11 12 13 14 15 16 17 ",13
db "18 19 20 21 22 23 24 ",13
db "25 26 27 28 29 30 31 ",0
color db 181
x db 0 ;SCREEN COORDINATE (COL).
y db 0 ;SCREEN COORDINATE (ROW).
.code
;INITIALIZE DATA SEGMENT.
mov ax,@data
mov ds,ax
;SWITCH SCREEN TO GRAPHICS MODE.
mov ah,0
mov al,13h ;320x240x256.
int 10H
mov di, offset jan
while:
call gotoxy ;SET CURSOR POSITION FOR CURRENT CHAR.
mov al, [ di ] ;CHAR TO DISPLAY.
cmp al, 13 ;IF CHAR == 13
je linebreak ;THEN JUMP TO LINEBREAK.
cmp al, 0 ;IF CHAR == 0
je finish ;THEN JUMP TO FINISH.
call char_display ;DISPLAY CHAR IN AL WITH "COLOR".
inc x ;NEXT CHARACTER GOES TO THE RIGHT.
jmp next_char
linebreak:
inc y ;MOVE TO NEXT LINE.
mov x, 0 ;X GOES TO THE LEFT.
next_char:
inc di ;NEXT CHAR IN "JAN".
jmp while
finish:
;WAIT FOR ANY KEY.
mov ah,7
int 21h
;FINISH THE PROGRAM PROPERLY.
mov ax,4c00h
int 21h
;-------------------------------------------------
;DISPLAY ONE CHARACTER IN "AL" WITH "COLOR".
proc char_display
mov ah, 9
mov bh, 0
mov bl, color ;ANY COLOR.
mov cx, 1 ;HOW MANY TIMES TO DISPLAY CHAR.
int 10h
ret
endp
;-------------------------------------------------
proc gotoxy
mov dl, x
mov dh, y
mov ah, 2 ;SERVICE TO SET CURSOR POSITION.
mov bh, 0 ;PAGE.
int 10h ;BIOS SCREEN SERVICES.
ret
endp
My graphics algorithm requires the use of char(13) for linebreaks (not 13,10) and the string to finish with 0.
Ghost
Code For All combinations of foreground and background colors:
Include Irvine32.inc
.data
str1 byte "F",0dh,0ah,0 ;character initialized (0dh,0ah for next line)
foreground Dword ? ;variable declaration
background Dword ?
counter Dword ?
.code
main PROC
mov ecx,16 ; initializing ecx with 16
l1: ;outer loop
mov counter,ecx
mov foreground,ecx
dec foreground
mov ecx,16
l2: ;inner loop
mov background , ecx
dec background
mov eax,background ; Set EAX = background
shl eax,4 ; Shift left, equivalent to multiplying EAX by 16
add eax,foreground ; Add foreground to EAX
call settextcolor ;set foreground and background color
mov edx, offset str1 ; string is moved to edx for writing
call writestring
loop l2 ;calling loop
mov ecx,counter
loop l1
exit
main ENDP
END main
Ghost
Include Irvine32.inc
.data
str1 BYTE "different color string",0dh,0ah,0 ;string initializing
counter dword ? ;save loop value in counter
.code
main PROC
mov ecx,4 ;loop repeated 4 times
l1:
mov counter,ecx ;counter save loop no
mov eax,counter ;eax contain loop no
;which is equal to counter
; color no
call SetTextColor ;Call color library
mov edx,offset str1 ;string reference is moved to edx
call WriteString ;string is write from reference
loop l1 ; calling loop l1
exit
main ENDP
END main
来源：https://stackoverflow.com/questions/29460318/how-to-print-colored-string-in-assembly-language