Experiment 3 Transfer Instruction Jump Principle and Simple Application Programming

IV. EXPERIMENTAL CONCLUSIONS

1. Experimental Task 1

Give the program task1.asm source code, and run a screenshot

 1 assume cs:code, ds:data
 2 
 3 data segment
 4     x db 1, 9, 3   ;x Initial value is 0 (offset address)
 5     len1 equ $ - x ; symbolic constants, $The offset address for the next data item, in this example, is 3
 6 
 7     y dw 1, 9, 3   ;y Is 3
 8     len2 equ $ - y ; symbolic constants, $The offset address for the next data item, in this example, is 9
 9 data ends
10 
11 code segment
12 start:
13     mov ax, data
14     mov ds, ax
15 
16     mov si, offset x ; Take Symbols x Corresponding offset address 0 -> si
17     mov cx, len1 ; From Symbol x Number of consecutive byte data items to start with -> cx
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

Answer Questions 1

(1) line27, the assembly instruction loop s1 jumps according to the amount of displacement. Through debug disassembly, look at its machine code and analyze how much displacement it jumps? (Displacement values are answered in decimal numbers) From the perspective of the CPU, how to calculate the offset address of the instruction following the jump label s1.

Machine code: F2 (h) --> 11110010 (b) --> 00001110 (b) --> -12 (d decimal) (complement --> source code)
Calculation of displacement: d(h) - 19(h) -->13-25= -12 (d decimal)
The CPU subtracts the current offset address from the target offset address and calculates the offset address of the instruction following the jump label s1.

Answer Questions 2

(2) line44, the assembly instruction loop s2 jumps according to the amount of displacement. Through debug disassembly, look at its machine code and analyze how much displacement it jumps? (Displacement values are answered in decimal numbers) From the perspective of the CPU, how to calculate the offset address of the instruction following the jump label s2.

  Displacement: F0(h)

Question 3

(3) Disassembly screenshots of debugging observations in debug when the above analysis is attached

  2. Experimental Task 2

Give program task2.asm source code

 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

  Given analysis, debugging, validation, register (ax) =? (bx) =? (cx) =? Attach a screenshot of the debug results interface.

(1) According to the jump principle of call instruction, theoretical analysis is made first. Register (ax) =? Register (bx) =? Register (cx) = before program execution (line31)?

Register (ax)=   Offset address of S1   Register (bx)=   Offset address of S2   Register (cx)=   Segment Address for S2

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

Agreement.

3. Experimental Task 3

Give program source task3.asm

 1 assume cs:code, ds:data
 2  
 3 data segment
 4     x    db  99, 72, 85, 63, 89, 97, 55
 5     len    equ $- x
 6 data ends
 7  
 8 code segment
 9     main:     
10         mov ax, data
11         mov ds, ax
12         
13         mov cx, len
14         mov si, offset x
15     print:
16         mov al, [si]
17         mov ah, 0
18         call printNumber
19         call printSpace
20         inc si
21         loop print
22  
23         mov ah, 4ch
24         int 21h
25  
26     ;Function: Output a two-digit number in decimal form
27     ;Entry parameter: register ax(Data to be output --> ax)
28     ;Export parameters: none
29     printNumber:
30         mov bl, 10
31         div bl
32         mov bx, ax
33         
34         mov ah, 2
35  
36         mov dl, bl    ; Printer(10 position)
37         or dl, 30h
38         int 21h
39  
40         mov dl, bh    ; Print Remainder(Bits)
41         or dl, 30h
42         int 21h
43         ret
44              
45     printSpace:
46         mov ah, 2
47         mov dl, ' '
48         int 21h
49         ret
50  
51 code ends
52 end main

Run Test Screenshot

4. Experimental Task 4

Give program source task4.asm

 1 assume cs:code
 2 data segment
 3     str db    'try'
 4     len equ    $ - str
 5 data ends
 6 code segment
 7     main:
 8         mov ax, 0b800h
 9         mov es, ax
10     
11     first_print:
12         mov ax, data
13         mov ds, ax
14         mov si, offset str
15         mov cx, len
16         mov bl, 00000010b    ; Green on Black
17         mov bh, 0            ; Line 0
18         call printStr
19     
20     second_print:
21         mov si, offset str
22         mov cx, len
23         mov bl, 00000100b    ; Red on Black
24         mov bh, 24            ; Line 24
25         call printStr
26     
27         mov ah, 4ch
28         int 21h
29  
30     ; Entry parameters:
31       ;    Address of first character in string --> ds:si(Where the segment address of the string is -> ds, Offset Address for String Start Address -> si)
32       ;    String Length --> cx
33       ;    String color --> bl
34       ;    Specify Row --> bh (Value:0 ~24)
35     printStr:
36         push bp        
37         push di
38  
39         mov ah, 0
40         mov al, 160
41         mul bh                    
42         mov bp, ax                    
43         mov di, si                    
44         printChar:
45             mov al, ds:[si]
46             mov es:[bp+di], al        ; character
47             mov es:[bp+di+1], bl    ; colour
48             inc si
49             inc di
50             inc di                    
51             loop printChar
52         
53         pop bp                        
54         pop di
55         ret 
56 code ends
57 end main

Run Test Screenshot

5. Experimental Task 5

Give program source task5.asm

 1 assume cs:code, ds:data
 2 data segment
 3     stu_no db '201983290357'
 4     len = $ - stu_no
 5 data ends
 6  
 7 code segment
 8     main:
 9         call print_blue_screen
10         call print_stu_no
11  
12         mov ah, 4ch
13         int 21h
14     print_blue_screen:
15         push ax        
16         push es
17         push si
18  
19         mov ax, 0b800h
20         mov es, ax
21         mov cx, 2000
22         mov si, 1
23         single_blue:
24             mov byte ptr es:[si], 00010000b
25             inc si
26             inc si
27             loop single_blue
28         
29         pop si        
30         pop es
31         pop ax        
32     ret
33     
34     print_stu_no:
35         push ax        
36         push es
37         push si
38         push ds
39         push di
40     prefix:
41         mov ax, 0b800h
42         mov es, ax
43         mov cx, 34
44         mov si, 3840    ; si Store offset address for each video output
45         call print_dash
46     content:
47         mov ax, data
48         mov ds, ax
49         mov cx, len
50         mov di, 0        ; di Deposit data Offset address for each character in
51         single_no:
52             mov al, ds:[di]
53             inc di
54             mov byte ptr es:[si], al 
55             inc si
56             mov byte ptr es:[si], 00010111b
57             inc si
58             loop single_no
59     postfix:
60         mov cx, 34
61         call print_dash
62         
63         pop di
64         pop ds
65         pop si
66         pop es
67         pop ax
68     ret
69     
70     ;    input parameter:
71     ;    Base address shown si
72     ;    Output Length cx
73     ;    output:
74     ;    Base address after iteration si
75     print_dash:
76         single_dash:
77             mov byte ptr es:[si], '-'
78             inc si
79             mov byte ptr es:[si], 00010111b
80             inc si
81             loop single_dash
82     ret
83  
84 code ends
85 end main

Run Test Screenshot

V. EXPERIMENTAL SUMMARY

1. Symbol constant, $refers to the offset address of the next data item.

2.EQU is an equivalent command:

COUNT   EQU   100; Make the COUNT value 100 and allocate 0 bytes to the variable in the memory;
COUNT DB 100; It should be an offset address with count value of 100.

3. The OFFSET operator returns the offset of the data label. This offset is in bytes and represents the distance between the data label and the starting address of the data segment.

Posted on Thu, 02 Dec 2021 12:28:01 -0500 by Frozenlight777