Experiment 3 transfer instruction jump principle and its simple application programming

 

Experimental task 1

Using any text editor, enter the 8086 assembler source code task1.asm.

task1.asm

 1 assume cs:code, ds:data
 2 
 3 data segment
 4     x db 1, 9, 3
 5     len1 equ $ - x
 6 
 7     y dw 1, 9, 3
 8     len2 equ $ - y
 9 data ends
10 
11 code segment
12 start:
13     mov ax, data
14     mov ds, ax
15 
16     mov si, offset x
17     mov cx, len1
18     mov ah, 2
19  s1:mov dl, [si]
20     or dl, 30h
21     int 21h
22 
23     mov dl, ' '
24     int 21h
25 
26     inc si
27     loop s1
28 
29     mov ah, 2
30     mov dl, 0ah
31     int 21h
32 
33     mov si, offset y
34     mov cx, len2/2
35     mov ah, 2
36  s2:mov dx, [si]
37     or dl, 30h
38     int 21h
39 
40     mov dl, ' '
41     int 21h
42 
43     add si, 2
44     loop s2
45 
46     mov ah, 4ch
47     int 21h
48 code ends
49 end start

 

question answering:

① line27, when the assembly instruction loop s1 jumps, it jumps according to the displacement. Check the machine code through debug disassembly and analyze the jump displacement? (the displacement value is answered in decimal) from the perspective of the CPU, explain how to calculate the offset address of the instruction after the jump label s1.

A: it can be seen from E2F2 that the displacement is F2 and the decimal is - 14, so the jump displacement is - 14. When loop is executed, the offset address of ip pointing to the next instruction is 00190019h + f2h, which is the offset address of label s1

 

② line44. When the assembly instruction loop s2 jumps, it jumps according to the displacement. Check the machine code through debug disassembly and analyze the jump displacement? (the displacement value is answered in decimal) from the perspective of the CPU, explain how to calculate the offset address of the instruction after the jump label s2.

       A: according to E2F0, the displacement is F0, and the decimal system is - 16,   Therefore, the jump displacement is - 16. When loop is executed, the offset address of ip pointing to the next instruction is 00390039h + F0H, which is the offset address of label s2

 

③ Attach the disassembly screenshot of debugging observation in debug during the above analysis

Screenshot:

 

 

Experimental task 2

Using any text editor, enter the 8086 assembler source code task2.asm.

task2.asm

 1 assume cs:code, ds:data
 2 
 3 data segment
 4     dw 200h, 0h, 230h, 0h
 5 data ends
 6 
 7 stack segment
 8     db 16 dup(0)
 9 stack ends
10 
11 code segment
12 start:  
13     mov ax, data
14     mov ds, ax
15 
16     mov word ptr ds:[0], offset s1
17     mov word ptr ds:[2], offset s2
18     mov ds:[4], cs
19 
20     mov ax, stack
21     mov ss, ax
22     mov sp, 16
23 
24     call word ptr ds:[0]
25 s1: pop ax
26 
27     call dword ptr ds:[2]
28 s2: pop bx
29     pop cx
30 
31     mov ah, 4ch
32     int 21h
33 code ends
34 end start

① According to the jump principle of call instruction, it is analyzed theoretically that before the program executes to exit (line31), register (ax) =? Register (bx) =? Register (cx) =?

Answer: (ax) = offset address of label s1, (bx) = offset address of label s2, (cx) is the value of cs register.

 

② Assemble and link the source program to get the executable program task2.exe. Use debug to observe and verify whether the debugging results are consistent with the theoretical analysis results.

 

The analysis is consistent with the results

  Experimental task 3

Source code:

 1 assume cs:code, ds:data
 2 data segment
 3     x db 99, 72, 85, 63, 89, 97, 55
 4     len equ $- x
 5 data ends
 6 
 7 code segment
 8 start:
 9     mov ax,data
10     mov ds,ax
11     mov si,offset x
12     mov cx,len
13 
14 s:
15     call printNumber
16     call printSpace
17     inc si
18     loop s
19     mov ah, 4ch 
20     int 21h
21 
22 printNumber:
23     mov ah,0
24     mov al,[si]
25     mov bl,10
26     div bl
27    
28    mov bx,ax
29    mov ah,2
30    mov dl,bl
31    or dl, 30h
32    int 21h
33   
34    mov ah, 2
35    mov dl, bh
36    or dl, 30h
37    int 21h
38    ret
39    
40  printSpace:
41    mov ah, 2
42    mov  dl, ' '
43    int 21h
44    ret 
45 
46 code ends
47 end start

 

result:

 

Same as expected

 

Experimental task 4

Source code:

 1 assume cs:code,ds:data
 2 
 3 data segment
 4     str db 'try'
 5     len equ $ - str
 6 data ends
 7 
 8 code segment
 9     start:
10           mov ax, data
11           mov ds, ax
12     
13           mov bl, 2
14           mov bh, 0
15           call printStr
16 
17           mov bl, 4    
18           mov bh, 24
19           call printStr
20 
21           mov ax, 4c00h
22           int 21h 
23 
24 printStr:
25     mov ax, 0b800h
26     mov es, ax
27 
28     mov si, offset str
29     mov cx, len
30 
31     mov al,0A0h
32     mul bh
33     mov di, ax
34     
35 s:  mov al,[si]
36     mov es:[di], al
37     mov es:[di+1], bl
38     inc si
39     add di, 2
40     loop s    
41     ret
42 
43 code ends
44 end start

 

 

  The results are the same as expected

 

Experimental task 5

Source code:

 1 assume cs:code, ds:data
 2 
 3 data segment 
 4     stu_no db '201983290272'
 5     len = $ - stu_no
 6 data ends
 7 
 8 code segment
 9 start:
10     mov ax, data
11     mov ds, ax
12      
13     mov ax, 0b800h    
14     mov es, ax
15     mov al, 24
16     mov dl, 80
17     mul dl
18     mov cx, ax
19     mov di, 0
20 
21 s1:mov al, 17h           ;Except for the last line, the page is set in blue
22     mov es:[di], byte ptr 20h
23     mov es:[di+1], al
24     add di, 2
25     loop s1
26     
27     mov ax, 00a0h
28     mov bh, 24
29     mul bh
30     mov di, ax
31     call printStr
32     
33     mov si, offset stu_no
34     mov cx, len
35     call printNum
36     
37     call printStr
38 
39     mov ah, 4ch
40     int 21h
41 
42 printStr:
43     mov ax, 80
44     sub ax, len
45     mov bl, 2
46     div bl
47     mov ch, 0
48     mov cl, al
49 
50 s2:mov ah, 17h
51     mov al, '-'
52     mov es:[di], ax
53     add di, 2
54     loop s2
55     ret
56    
57 printNum:
58 s3:mov ah, 17h
59     mov al, [si]
60     mov es:[di], ax
61     inc si
62     add di, 2
63     loop s3
64     ret
65 
66 code ends
67 end start


The screenshot of the result is as follows

 

 

The results are the same as expected

 

Posted on Thu, 02 Dec 2021 17:58:10 -0500 by tsapat