『第 57 楼』:
使用 LLM 解释/回答一下
请看罗云彬的源码例:
.286
CODE SEGMENT
ASSUME CS:CODE,DS:CODE
ORG 100H
START:
jmp install
; ........ 0000 0000 00h
; ..xxxxxx 0011 1111 3fh
; .x...... 0100 0000 40h
; .x.xxxxx 0101 1111 5fh
; .x.x..x. 0101 0010 52h
; .xxx.x.. 0111 0100 74h
; ....x.xx 0000 1011 0bh
; ...x.x.. 0001 0100 14h
; ..x.x..x 0010 1001 29h
; .x..x.x. 0100 1010 4ah
; .x..x..x 0100 1001 49h
; .x...x.. 0100 0100 44h
; .x....xx 0100 0011 43h
; ..xxxxxx 0011 1111 3fh
; ........ 0000 0000 00h
; ........ 0000 0000 00h
CHAR db 00h,3fh,40h,5fh,52h,74h,0bh,14h ;left part of phone
db 29h,4ah,49h,44h,43h,3fh,00h,00h
db 00h,0f8h,04h,0f4h,094h,5ch,0a0h,50h ;right part of phone
db 28h,0a4h,24h,44h,84h,0f8h,00h,00h
STRn db 'This is a phone:',0ceh,0cfh,'$'
install:
mov ax,1100h ;功能号
mov bx,1000h ;BH=10H,每字符点阵为16字节
mov cx,2 ;共要装入2字节
mov dx,00ceh ;从00CEH开始(00CEH,00CFH)
push ds
pop es
mov bp,offset char ;ES:BP为字符点阵数据地址
int 10h
mov ah,9 ;把字符 ce,cf显示出来看看,是什么!
mov dx,offset strn
int 21h
ret
CODE ENDS
END START
Please provide the specific part that needs to be translated. If you want the entire content translated, here it is:
Please look at the source code example of Luo Yunbin:
.286
CODE SEGMENT
ASSUME CS:CODE,DS:CODE
ORG 100H
START:
jmp install
; ........ 0000 0000 00h
; ..xxxxxx 0011 1111 3fh
; .x...... 0100 0000 40h
; .x.xxxxx 0101 1111 5fh
; .x.x..x. 0101 0010 52h
; .xxx.x.. 0111 0100 74h
; ....x.xx 0000 1011 0bh
; ...x.x.. 0001 0100 14h
; ..x.x..x 0010 1001 29h
; .x..x.x. 0100 1010 4ah
; .x..x..x 0100 1001 49h
; .x...x.. 0100 0100 44h
; .x....xx 0100 0011 43h
; ..xxxxxx 0011 1111 3fh
; ........ 0000 0000 00h
; ........ 0000 0000 00h
CHAR db 00h,3fh,40h,5fh,52h,74h,0bh,14h ;left part of phone
db 29h,4ah,49h,44h,43h,3fh,00h,00h
db 00h,0f8h,04h,0f4h,094h,5ch,0a0h,50h ;right part of phone
db 28h,0a4h,24h,44h,84h,0f8h,00h,00h
STRn db 'This is a phone:',0ceh,0cfh,'$'
install:
mov ax,1100h ;Function number
mov bx,1000h ;BH=10H, each character dot matrix is 16 bytes
mov cx,2 ;Total 2 bytes to load
mov dx,00ceh ;Start from 00CEH (00CEH, 00CFH)
push ds
pop es
mov bp,offset char ;ES:BP is the character dot matrix data address
int 10h
mov ah,9 ;Display characters ce, cf to see what it is!
mov dx,offset strn
int 21h
ret
CODE ENDS
END START
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