# Assembly language course - Experiment 2 - Compilation and debugging of assembly source program of multiple logic segments

## Experiment 1   Compilation and debugging of assembly source program for multiple logic segments

``` 1 assume ds:data, cs:code, ss:stack
2
3 data segment
4     db 16 dup(0)
5 data ends
6
7 stack segment
8     db 16 dup(0)
9 stack ends
10 code segment
11 start:
12     mov ax, data
13     mov ds, ax
14
15     mov ax, stack
16     mov ss, ax
17     mov sp, 16
18
19     mov ah, 4ch
20     int 21h
21 code ends
22 end start```
• task1_1. Screenshot from the end of debugging to line17 and before line19:

① In debug, execute until the end of line17 and before line19. Record this time: register (DS) = 076A, register (SS) = 076B, register (CS)=   076C
② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X-2 and the segment address of the stack is X-1

``` 1 assume ds:data, cs:code, ss:stack
2
3 data segment
4     db 4 dup(0)
5 data ends
6
7 stack segment
8     db 8 dup(0)
9 stack ends
10 code segment
11 start:
12     mov ax, data
13     mov ds, ax
14
15     mov ax, stack
16     mov ss, ax
17     mov sp, 8
18
19     mov ah, 4ch
20     int 21h
21 code ends
22 end start```
• task1_2 screenshot of register DS, CS and SS values before the end of line17 and line19:

① In debug, execute until the end of line17 and before line19. Record this time: register (DS) = 076A, register (SS) = 076B, register (CS) = 076C
② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X-2 and the segment address of the stack is X-1

``` 1 assume ds:data, cs:code, ss:stack
2
3 data segment
4     db 20 dup(0)
5 data ends
6
7 stack segment
8     db 20 dup(0)
9 stack ends
10 code segment
11 start:
12     mov ax, data
13     mov ds, ax
14
15     mov ax, stack
16     mov ss, ax
17     mov sp, 20
18
19     mov ah, 4ch
20     int 21h
21 code ends
22 end start```
• task1_3 screenshot of register DS, CS and SS values before the end of line17 and line19:

① In debug, execute until the end of line17 and before line19. Record this time: register (DS) = 076A, register (SS) = 076C, register (CS) = 076E
② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X-4 and the segment address of the stack is X-2

``` 1 assume ds:data, cs:code, ss:stack
2 code segment
3 start:
4     mov ax, data
5     mov ds, ax
6
7     mov ax, stack
8     mov ss, ax
9     mov sp, 20
10
11     mov ah, 4ch
12     int 21h
13 code ends
14
15 data segment
16     db 20 dup(0)
17 data ends
18
19 stack segment
20     db 20 dup(0)
21 stack ends
22 end start```
• task1_4 screenshot of register DS, CS and SS values before the end of line17 and line19:

① In debug, execute until the end of line9 and before line11. Record this time: register (DS) = 076C, register (SS) = 076E, register (CS) = 076A
② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X+2 and the segment address of the stack is X+4

Based on the practice and observation of the above four experimental tasks, summarize and answer:

① for the segment defined below, after the program is loaded, the actual memory space allocated to the segment is 16*Math.ceil(N/16).

```1 xxx segment
2     db N dup(0)
3 xxx ends```

② if the program Task1_ 1.asm, task1_ 2.asm, task1_ 3.asm, task1_ 4. In ASM, if the pseudo instruction end start is changed to end, which program can still be executed correctly. The reasons are analyzed and explained in combination with the conclusions obtained from practical observation.

A: only task1_4.asm can be executed correctly. Observed through practice: Task1_ 1~task1_ cs in 3 has no code that needs to be executed. Without end start, it means that the end position of code execution is not defined in the program. In Task1_ 1~task1_ In 3, the data segment is executed and the code segment execution fails. In task1_4, because the code segment is defined at the beginning, it is called task1_4 can be executed successfully.

• Assembly source code:
``` 1 assume cs:code
2 code segment
3 start:
4     mov ax,0b800h
5
6     mov ds,ax
7     mov bx,0f00h
8     mov ax,0403h
9
10     mov cx,80
11     s:
12     mov ds:[bx],ax
14     loop s
15
16     mov ah, 4ch
17     int 21h
18 code ends
19 end start```
• Screenshot of operation results:

• Complete assembly source code:
``` 1 assume cs:code
2 data1 segment
3     db 50, 48, 50, 50, 0, 48, 49, 0, 48, 49 ; ten numbers
4 data1 ends
5
6 data2 segment
7     db 0, 0, 0, 0, 47, 0, 0, 47, 0, 0       ; ten numbers
8 data2 ends
9
10 data3 segment
11     db 16 dup(0)
12 data3 ends
13
14 code segment
15 start:
16     mov ax, data1
17     mov ds,ax
18     mov bx,0
19     mov cx,10h
20     s:
21     mov ax,ds:[bx]
23     mov ds:[bx+20h],ax
24     inc bx
25     loop s
26
27     mov ah, 4ch
28     int 21h
29 code ends
30 end start```

• debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3 before data items are added in turn:

• After adding in sequence, the debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3:

•   Complete assembly source code:
``` 1 assume cs:code
2
3 data1 segment
4     dw 2, 0, 4, 9, 2, 0, 1, 9
5 data1 ends
6
7 data2 segment
8     dw 8 dup(?)
9 data2 ends
10
11 code segment
12 start:
13     mov ax,data1
14     mov ds,ax
15     mov ax,data2
16     mov ss,ax
17     mov sp,16
18     mov bx,0
19     mov cx,8
20 s :push ds:[bx]
22      loop s
23
24     mov ah, 4ch
25     int 21h
26 code ends
27 end start```

•   Before the program exits, use the d command to view a screenshot of the memory space corresponding to data segment data2.

``` 1 assume cs:code, ds:data
2 data segment
3         db 'Nuist'
4         db 2,3,4,5,6
5 data ends
6
7 code segment
8 start:
9         mov ax, data
10         mov ds, ax
11
12         mov ax, 0b800H
13         mov es, ax
14
15         mov cx, 5
16         mov si, 0
17         mov di, 0f00h
18 s:      mov al, [si]
19         and al, 0dfh
20         mov es:[di], al
21         mov al, [5+si]
22         mov es:[di+1], al
23         inc si
25         loop s
26
27         mov ah, 4ch
28         int 21h
29 code ends
30 end start```
• Screenshot of operation results:

• Use the debug tool to debug the program, and use the g command to execute it once before the program returns (i.e. after ine25 and before line27):

• What is the function of line19 in the source code?

A: the code of line19 is: and al, 0dfh, where and means to perform and operate by bit, that is, AL performs and operates on DF(1101 1111), that is, set the third position on the left as 0

• What is the purpose of the byte data in the data segment line4 in the source code?

A: the purpose of db is to allocate a storage unit for the following numbers.

``` 1 assume cs:code, ds:data
2
3 data segment
4     db 'Pink Floyd      '
5     db 'JOAN Baez       '
6     db 'NEIL Young      '
7     db 'Joan Lennon     '
8 data ends
9
10 code segment
11 start:
12    mov ax,data
13    mov ds,ax
14    mov bx,0
15    mov cx,4
16 s: or [bx],byte ptr 00100000b
18    loop s
19
20    mov ah, 4ch
21    int 21h
22 code ends
23 end start```

• Before the program exits, use the d command to view a screenshot of the memory space corresponding to the data segment data:

``` 1 assume cs:code, ds:data, es:table
2
3 data segment
4     db '1975', '1976', '1977', '1978', '1979'
5     dw  16, 22, 382, 1356, 2390
6     dw  3, 7, 9, 13, 28
7 data ends
8
9 table segment
10     db 5 dup( 16 dup(' ') )  ;
11 table ends
12
13 code segment
14 start:
15     mov ax,data
16     mov ds,ax
17     mov ax,table
18     mov es,ax
19
20     mov cx,5
21     mov si,0
22     mov bx,0
23 s:  mov ax,[si]
24     mov es:[bx],ax
25     mov ax,[si+2]
26     mov es:[bx+2],ax
29     loop s   ;particular year
30
31     mov cx,5
32     mov si,20
33     mov bx,5
34 s1:mov ax,[si]
35     mov es:[bx],ax
36     mov ax,0
37     mov es:[bx+2],ax
40     loop s1;  ;income
41
42     mov cx,5
43     mov si,30
44     mov bx,10
45 s2:mov ax,[si]
46     mov es:[bx],ax
49     loop s2   ;Number of employees
50
51     mov cx,5
52     mov si,5
53 s3:mov ax,es:[si]
54     mov bl,es:[si+5]
55     div bl
56     mov es:[si+8],al
58     loop s3   ;Per capita income
59
60     mov ah, 4ch
61     int 21h
62 code ends
63 end start```
• Commissioning screenshot:

• View the screenshot of the original data information of the table segment:

• Run in debug until the program exits, use the d command to view the screenshot of the memory space corresponding to the table segment, and confirm whether the information is structurally written to the specified memory as required

Added by hemlata on Tue, 09 Nov 2021 12:04:18 +0200