微机原理课程设计_打字计时练习_汇编课程设计|精品课程网站设计|课程设计网报告总结心得
说明:这是一个打字计时练习的程序,在缓冲区中预放了一些字母,运行时,可按照屏幕上
显示的字母输入练习,每输入完一行按回车键后,可显示出练习输入的时间.
stack segment para stack 'stack'
db 256 dup(0)
top label word
stack ends
data segment para public 'data'
buffer db 16h dup(0)
bufpt1 dw 0
bufpt2 dw 0
kbflag db 0
prompt db ' *please practise typing*',0dh,0ah,'$'
scantab db 0,0,'1234567890-=',8,0
db 'qwertyuiop[]',0dh,0
db 'asdfghjkl;',0,0,0,0
db 'zxcvbnm,./',0,0,0
db ' ',0,0,0,0,0,0,0,0,0,0,0,0,0
db '789-456+1230.'
even
oldcs9 dw ?
oldip9 dw ?
str1 db 'abcd efgh ijkl mnop qrst uvwx yz.'
db 0dh,0ah,'$'
str2 db 'there are some newspapers on the table.'
db 0dh,0ah,'$'
str3 db 'there are some clouds in the sky.'
db 0dh,0ah,'$'
str4 db 'she always eats her lunch at noon.'
db 0dh,0ah,'$'
str5 db 'i do not like autumn and winter.'
crlf db 0dh,0ah,'$'
colon db ':','$'
even
saddr dw str1,str2,str3,str4,str5
count dw 0
sec dw 0
min dw 0
hours dw 0
save_lc dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data,es:data,ss:stack
main proc far
start:
mov ax,stack
mov ss,ax
mov sp,offset top
push ds
sub ax,ax
push ax
mov ax,data
mov ds,ax
mov es,ax
; mov ah,0
; mov al,4
; int 21h
; mov ah,0bh
; mov bh,0
; mov bl,4
; int 11h
mov ah,35h
mov al,09h
int 21h
mov oldcs9,es
mov oldip9,bx
push ds
mov dx,seg kbint
mov ds,dx
mov dx,offset kbint
mov al,09h
mov ah,25h
int 21h
pop ds
mov ah,35h
mov al,1ch
int 21h
mov save_lc,bx
mov save_lc+2,es
push ds
mov dx,seg clint
mov ds,dx
mov dx,offset clint
mov al,1ch
mov ah,25h
int 21h
pop ds
in al,21h
and al,11111100b
out 21h,al
first: mov ah,0
mov al,3
int 10h
mov dx,offset prompt
mov ah,9
int 21h
mov si,0
next: mov dx,saddr[si]
mov ah,09h
int 21h
mov count,0
mov sec,0
mov min,0
mov hours,0
sti
forever:
call kbget
test kbflag,80h
jnz endint
push ax
call dispchar
pop ax
cmp al,0dh
jnz forever
mov al,0ah
call dispchar
call disptime
lea dx,crlf
mov ah,09h
int 21h
add si,2
cmp si,5*2
jne next
jmp first
endint: cli
push ds
mov dx,save_lc
mov ax,save_lc+2
mov ds,ax
mov al,1ch
mov ah,25h
int 21h
pop ds
push ds
mov dx,oldip9
mov ax,oldcs9
mov ds,ax
mov al,09h
mov ah,25h
int 21h
pop ds
sti
ret
main endp
clint proc near
push ds
mov bx,data
mov ds,bx
lea bx,count
inc word ptr[bx]
cmp word ptr[bx],18
jne return
call inct
adj:
cmp hours,12
jle return
sub hours,12
return:
pop ds
sti
iret
clint endp
inct proc near
mov word ptr[bx],0
add bx,2
inc word ptr[bx]
cmp word ptr[bx],60
jne exit
call inct
exit: ret
inct endp
disptime proc near
mov ax,min
call bindec
mov bx,0
mov al,':'
mov ah,0eh
int 10h
mov ax,sec
call bindec
mov bx,0
mov al,':'
mov ah,0eh
int 10h
mov bx,count
mov al,55d
mul bl
call bindec
ret
disptime endp
bindec proc near
mov cx,100d
call decdiv
mov cx,10d
call decdiv
mov cx,1
call decdiv
ret
bindec endp
decdiv proc near
mov dx,0
div cx
mov bx,0
add al,30h
mov ah,0eh
int 10h
mov ax,dx
ret
decdiv endp
kbget proc near
push bx
cli
mov bx,bufpt1
cmp bx,bufpt2
jnz kbget2
cmp kbflag,0
jnz kbget3
sti
pop bx
jmp kbget
kbget2:
mov al,[buffer+bx]
inc bx
cmp bx,16h
jc kbget3
mov bx,0
kbget3:
mov bufpt1,bx
pop bx
ret
kbget endp
kbint proc near
push bx
push ax
in al,60h
push ax
in al,61h
or al,80h
out 61h,al
and al,7fh
out 61h,al
pop ax
test al,80h
jnz kbint2
mov bx,offset scantab
xlat scantab
cmp al,0
jnz kbint4
mov kbflag,80h
jmp kbint2
kbint4:
mov bx,bufpt2
mov [buffer+bx],al
inc bx
cmp bx,16h
jc kbint3
mov bx,0
kbint3:
cmp bx,bufpt1
jz kbint2
mov bufpt2,bx
kbint2: cli
mov al,20h
out 20h,al
pop ax
pop bx
sti
iret
kbint endp
dispchar proc near
push bx
mov bx,0
mov ah,0eh
int 10h
pop bx
ret
dispchar endp
code ends
end start