蒙珣的博客

活好当下，做好今天该做的事情。

0%

实验15 安装新的int9中断例程

Posted on Edited on In
Symbols count in article: 3.5k Reading time ≈ 3 mins.

外中断不可屏蔽中断，中断类型码为2

1
2
3
4
5
pushf
IF = 0  TF = 0
push cs
push ip
ip = 2*4 cs = 2*4+2

外中断可屏蔽中断，和内中断一样

1
2
3
4
5
6
;1.获取中断类型码
;2.pushf
;3.IF = 0 TF = 0
push cs
push ip
IP = n*4    cs = n*4+2

编程：在屏幕中间依次显示”a-z”,并可以让人看清，在显示过程中，按下esc键后，改变显示颜色,依次显示”a-z”

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
assume cs:code
code segment

start:  mov ax,0b800h
   mov es,ax
   mov ah,'a'
   
  s: mov es:[160*12+40*2],ah
   inc ah
   cmp ah,'z'
   jna s
   mov ax,4c00h
   int 21h
   
code ends
end start

每显示一个字母后，延时一段时间，让人看清，再显示下一个字母
我们可以让cpu执行一段时间的空循环，cpu执行的速度非常块，所以循环次数一定要到
用两个16位寄存器存放32位的循环次数

1
2
3
4
5
6
7
8
      mov dx,10h
      mov ax,0
s:    sub ax,1
      sbb dx,0  ;带借位减法
      cmp ax,0
      jne s
      cmp dx,0
      jne s

上面的程序，循环1000000h次，我们可以将循环延时的程序写作一个子程序

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
assume cs:code
stack segment
 db 128 dup (0)
stack ends

code segment

start:      mov ax,stack
            mov ss,ax
            mov sp,128
            
            mov ax,0b800h
            mov es,ax
            mov ah,'a'
   s:       mov es:[160*12+40*2],ah
            call delay
            inc ah
            cmp ah,'z'
            jna s
            
            mov ax,4c00h
            int 21h
    
  delay:    push ax
            push dx
            mov dx,1000h  ;循环100000000h
            mov ax,0
    s1:     sub ax,1
            sbb dx,0
            cmp ax,0
            jne s1
            cmp dx,0
            jne s1
            pop dx
            pop ax
            ret
            
        code ends

        end start

键盘输入的处理过程:
1.键盘产生扫描码
2.扫描码送入60h端口
3.引发9号中断
4.cpuz执行int9中断例程处理键盘输入

键盘输入到达60h端口后，就会引发9好中断，cpu则转去执行int 9中断例程

1.从60h端口读出键盘输入
2.调用bios的int 9中断例程，处理其他硬件细节
3.判断是否为Esc的扫描码，如果是，改变显示的颜色后返回；如果不是则直接返回

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
assume cs:code

stack segment
    db 128 dup (0)
stack ends

data segment
    dw 0,0
data ends

code segment

start:      mov ax,stack
            mov ss,ax
            mov sp,128

            mov ax,data
            mov ds,ax

            mov ax,0
            mov es,ax

            ;将原来的int 9中断例程的入口地址保存在ds:0、ds:2单元中
            push es:[9*4]
            pop ds:[0]
            push es:[9*4+2]
            pop ds:[2]      

            ;在中断向量表中设置新的int 9中断例程的入口地址
            mov word ptr es:[9*4],offset int9
            mov es:[9*4+2],cs

            mov ax,0b800h
            mov es,ax
            mov ah,'a'
s:          mov es:[160*12+40*2],ah
            call delay
            inc ah
            cmp ah,'z'
            jna s

            mov ax,0
            mov es,ax

            ;将中断向量表中int 9中断例程的入口恢复为原来的地址
            push ds:[0] pop es:[9*4]
            push ds:[2]
            pop es:[9*4+2]

            mov ax,4c00h
            int 21h

delay:      push ax
            push dx

            mov dx,10h
            mov ax,0

s1:         sub ax,1
            sbb dx,0
            cmp ax,0
            jne s1
            cmp dx,0
            jne s1
            
            pop dx
            pop ax
            ret

;=======================================
int9:       push ax
            push bx
            push es

            int al,60h

            pushf
            pushf
            pop bx
            and bh,11111100b
            push bx
            popf
            
            ;对int指令进行模拟，调用原来的int9中断例程
            call dword ptr ds:[0]

            cmp al,1
            jne int9ret

            mov ax,0b800h
            mov es,ax

            ;将属性加1,改变颜色
            inc byte ptr es:[160*12+40*2+1]

int9ret:    pop es
            pop bx
            pop ax
            iret

code ends
end start

实验15 安装新的int9中断例程

安装一个新的 int9 中断例程，功能：在DOS下，按下”A”键后，除非不再松开，如果松开，就显示满屏幕的”A”，其他的按键照常处理

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
assume cs:code

stack segment
    db 128 dup (0)
stack ends

code segment

start:      mov ax,stack
            mov ss,ax
            mov sp,128

            call cpy_new_int9
            call sav_old_int9
            call set_new_int9

testA:      mov ax,1000H
            jmp testA

            mov ax,4c00h
            int 21h

;====================================
new_int9:   push ax

            in al,60h
            pushf
            call dword ptr cs:[200h]

            cmp al,9Eh
            jne int9Ret
            call set_screen_letter

int9Ret:    pop ax
            iret

;=======================================
set_screen_letter:
            push bx
            push cx
            push dx
            push es

            mov bx,0b800h
            mov es,bx
            mov bx,0
            mov dl,'A'
            mov cx,2000

setScreenLetter:
            mov es:[bx],dl
            add bx,2
            loop setScreenLetter

            pop es
            pop dx
            pop cx
            pop bx
            ret

new_int9_end:
            nop

;====================================
set_new_int9:
            mov bx,0
            mov es,bx

            cli
            mov word ptr es:[9*4],7E00H
            mov word ptr es:[9*4+2],0
            sti

            ret
            
;====================================
sav_old_int9:
            mov bx,0
            mov es,bx

            cli
            push es:[9*4]
            pop es:[200h]
            push es:[9*4+2]
            pop es:[202h]
            sti
            ret

;====================================
cpy_new_int9:
            mov bx,cs
            mov ds,bx
            mov si,offset new_int9

            mov bx,0
            mov es,bx
            mov di,7E00H

            mov cx,offset new_int9_end - offset new_int9
            cld
            rep movsb

            ret


;====================================

code ends

end start