#include <studio.h>
#include <dos.h>
void interrupt (*int9save) (void)
void interrupt eliminate_multiple_press(void)
{
int9save=getvect(9);
setvect(9,eliminate_multiple_press);
asm {
MOV AH,0
int 16h
MOV scan_temp,AH
CMP ZF,0
}
}
void interrupt uneliminate_multiple_press()
{
setvect(9,int9save);
}
void main(void)
{
char str[100];
int check=1;
char scan_temp;
unsigned int scan_code ;
eliminate_multiple_press();
printf("enter a word\n");
scanf("%s",str);
scan_code=(unsigned int) scan_temp;
printf("the word is:\n");
printf("%s",str);
uneliminate_multiple_press();
return ;
}
嘿,我正在写一个汇编代码..我正在尝试解决一个中断问题,它要求我做一个长按钮对待就像一个按钮,我被困在这里!所以请有人可以帮助我或给我一个方向如何继续...当我按下按钮ZF == 0当我离开按钮ZF == 1这可能有帮助
非常感谢
答案 0 :(得分:2)
在你自己的键盘中断处理程序中你可以有这样的东西(这是YASM / NASM语法,没有经过测试):
编辑:重写了代码,添加了评论和链接。
@my_int9:
cli
push ax ; push ax (you can create a stack frame too, if you wish).
push bx ; push bx.
in al,0x60 ; read scancode from keyboard port 60h to al.
cmp al,[cs:last_scancode] ; compare the current and last scancodes.
je @ready ; jump if its the same scancode, nothing to do.
test al,0x80 ; test highest bit of al to see if it's release
; or not.
; test does logical AND without saving the result,
; it only updates the flags
; (and al,0x80 would be OK too).
jnz @key_released ; jump if it's a released key.
; OK, we have a new key.
@new_key:
movzx bx,al ; copy the scancode from al to bx.
mov [cs:last_scancode],al ; store the current scancode into memory.
; Do something with the new key here.
; This is an example.
mov al,1
mov [cs:bx+keys_pressed],al ; set the corresponding byte of array to 1
; (pressed).
mov [cs:something_to_do],al ; set the flag "something to do" to 1.
; (so that the main code needs to scan through
; keys_pressed array only when there's at least
; 1 key that hasn't been handled yet).
@key_released:
; Do something here upon the key release if you wish...
; This really depends on what do you want to if with released keys.
; If you want that keypresses are handled even after the corresponding key
; is are released and that the the key release shouldn't cause any action,
; (in the case you don't poll that repeatedly), don't do anything here.
;
; If you want that keypresses are _not_ handled after the release, then set
; the corresponding byte of keys_pressed array to 0
; (uncomment the 3 lines below):
;
; and al,0x7F ; clear the highest bit.
; movzx bx,al ; copy the scancode from al to bx.
; mov [cs:bx+keys_pressed],byte 0 ; mark the key as not pressed.
@ready:
mov al,0x20 ; write byte 20h to port 20h to inform PIC.
out 0x20,al ; (programmable interrupt controller) that it's OK
; to continue sending interrupts.
pop bx
pop ax
sti
iret
last_scancode db 0
keys_to_handle db 0 ; in the main code you can poll for this.
; After handling the keys, set this to 0.
keys_pressed times 128 db 0 ; db 128 dup 0 in some other assemblers
; In the main program code scan through this if
; keys_to_handle is not zero, and set the
; corresponding byte to 0 to not handle it twice.
主代码(不属于中断控制器):
@main_code_loop:
test [cs:keys_to_handle], byte 0xFF ; check if there are keys to handle.
jz @no_keys_to_handle ; no, nothing to do.
; here scan through entire keys_pressed array and set handled keys'
; corresponding bytes to 0.
; remember to loop through the entire array, there can be several keys to handle.
mov [cs:keys_to_handle], byte 0 ; set keys_to_handle byte to 0.
@no_keys_to_handle:
; do something else
一些有用的链接:
OSDev: Interrupts:关于中断的有用文章。
OSDev: "8042" PS/2 Controller:有关键盘处理的有用信息。
OSDev: 8259 PIC:有关8259可编程中断控制器的信息。