实验4 8086标志寄存器及中断

实验1

【源码】

 1 assume cs:code, ds:data
 2 
 3 data segment
 4    x dw 1020h, 2240h, 9522h, 5060h, 3359h, 6652h, 2530h, 7031h
 5    y dw 3210h, 5510h, 6066h, 5121h, 8801h, 6210h, 7119h, 3912h
 6 data ends
 7 code segment 
 8 start:
 9     mov ax, data
10     mov ds, ax
11     mov si, offset x
12     mov di, offset y
13     call add128
14 
15     mov ah, 4ch
16     int 21h
17 
18 add128:
19     push ax
20     push cx
21     push si
22     push di
23 
24     sub ax, ax
25 
26     mov cx, 8
27 s:  mov ax, [si]
28     adc ax, [di]
29     mov [si], ax
30 add si,2
31 add di,2
32 
33 
34     loop s
35 
36     pop di
37     pop si
38     pop cx
39     pop ax
40     ret
41 code ends
42 end start

View Code

不可以，因为add指令会改变标志寄存器的值

前两行在加前保存了两个数据段的值，加后结果保存在了前者。

实验2

【源码】

 1 assume cs:code, ds:data
 2 data segment
 3         str db 80 dup(?)
 4 data ends
 5 
 6 code segment
 7 start:  
 8         mov ax, data
 9         mov ds, ax
10         mov si, 0
11 s1:        
12         mov ah, 1
13         int 21h
14         mov [si], al
15         cmp al, '#'
16         je next
17         inc si
18         jmp s1
19 next:
20         mov ah, 2
21         mov dl, 0ah
22         int 21h
23         
24         mov cx, si
25         mov si, 0
26 s2:    mov ah, 2
27         mov dl, [si]
28         int 21h
29         inc si
30         loop s2
31 
32         mov ah, 4ch
33         int 21h
34 code ends
35 end start

View Code

效果是以#号作为结束符，输出前面的所有内容

11-18行，用cmp比较输入字母，je表示当等于“#”时跳转到输出函数结束循环，否则继续循环读入内容

20-22行起换行功能，效果同实验三任务1

24-30行是对前面读入的内容进行输出。si先记录了字符串长度，赋值给cx作为循环次数。

任务3

【源码】

 1 assume ds:data, cs:code
 2 
 3 data segment
 4     x dw 91, 792, 8536, 65535, 2021
 5     len equ $ - x
 6 data ends
 7 
 8 code segment
 9 start:
10     mov ax, data
11     mov ds, ax
12     mov si,offset x        ; 取符号x对应的偏移地址0 -> si
13     mov cx,len/2        ;从符号x开始的连续字节数据项个数的一半 -> cx
14 
15     s:
16         mov ax,ds:[si]    
17         inc si
18         inc si                
19 
20         push cx          ;cx入栈防止计数器冲突       
21 
22         call printNumber    
23         call printSpace     
24 
25         pop cx              
26     loop s
27     
28     mov ah, 4ch
29     int 21h
30 
31 printNumber:
32     mov bx, 0     ;入栈计数器           
33 
34     s1:    
35         mov di, 10
36         mov dx,  0   ;dx置0            
37         div di        
38 
39         push dx         
40         inc bx          
41         
42         cmp ax,0     ;检测商是否为0 
43         jne s1       ;不等于则转移
44 
45 
46     mov cx, bx
47        
48 
49     s2:        ；输出数字
50         pop dx         
51         add dl,30h      
52         mov ah, 2       
53         int 21h
54         loop s2
55     ret
56 
57 
58 printSpace:      ;输出空格和上次实验一样
59     mov ah, 2
60     mov dl, 20h
61     int 21h
62     ret
63 
64 code ends
65 end start

View Code

任务4

【源码】

 1 assume cs:code,ds:data
 2 data segment
 3     str db "assembly language, it's not difficult but tedious"
 4     len equ $ - str
 5 data ends
 6 
 7 stack segment
 8     db 8 dup(?)
 9 stack ends
10 
11 code segment
12 start: 
13     mov ax, data
14     mov ds, ax
15     mov si,offset str        ; 取符号str对应的偏移地址0 -> si
16     mov cx,len        ;从符号str开始的连续字节数据项个数 -> cx
17    
18     call strupr     
19     call printStr   
20 
21     mov ax, 4c00h
22     int 21h
23 
24 
25 strupr:
26     push cx    ；cx入栈防止冲突
27     push si        
28 capital:
29         mov al, ds:[si] 
30         cmp al, 97        ;与a的ascii码比较
31         jb continue        ;小于则跳转
32         cmp al, 122     
33         ja continue     
34         and al, 0dfh    ;转换大写
35         mov ds:[si], al 
36     continue:
37         inc si          
38         loop capital
39 
40     pop si
41     pop cx          
42     ret             
43 
44 
45 printStr:
46     push cx         
47     push si
48 
49     print:          
50         mov ah, 2
51         mov dl, ds:[si]
52         int 21h
53         inc si
54     loop print
55 
56     pop si          
57     pop cx
58     ret            
59 
60 code ends
61 end start

View Code

任务5

【源码】

 1 assume cs:code, ds:data
 2 
 3 data segment
 4     str1 db "yes", '$'
 5     str2 db "no", '$'
 6 data ends
 7 
 8 code segment
 9 start:
10     mov ax, data
11     mov ds, ax
12 
13     mov ah, 1
14     int 21h
15 
16     mov ah, 2
17     mov bh, 0
18     mov dh, 24
19     mov dl, 70
20     int 10h
21 
22     cmp al, '7'
23     je s1
24     mov ah, 9
25     mov dx, offset str2
26     int 21h
27 
28     jmp over
29 
30 s1: mov ah, 9
31     mov dx, offset str1
32     int 21h
33 over:  
34     mov ah, 4ch
35     int 21h
36 code ends
37 end start