# 1, Experimental conclusion

•  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 before the end of line17 and line19
• ① In debug, execute until the end of line17 and before line19, and 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   X-2h, the segment address of stack is     X-1h   .
•  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. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values
• ① In debug, execute until the end of line17 and before line19, and record this time: register (DS)=     076A   ， Register (SS)=    076B  ， Register (CS)=     076C
• ② Suppose that after the program is loaded, the segment address of the code segment is x, and the segment address of the data segment is X    X-2h  ， The segment address of stack is     X-1h   .
•  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 the values of registers DS, CS and SS before the end of debugging to line17 and line19
• ① In debug, execute until the end of line17 and before line19, and 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     X-4h   ， The segment address of stack is   X-2h
•  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. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values

• ① In debug, execute until the end of line9 and before line11, and record this time: register (DS)=   076C   ， Register (SS)=     076E  ， Register (CS)=     076A
• ② Suppose that after the program is loaded, the segment address of the code segment is x, and the segment address of the data segment is X   X+2h   ， The segment address of stack is   X+4h  .
• 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     N/16 is rounded up and multiplied by 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.
• Answer: task1_4.asm can operate normally. end start not only informs the compiler of the end of the program, but also informs the compiler of where the entry of the program is. If the end is not used to indicate the program start label, the program runs from DS+10h at the beginning of the program by default. As shown in the figure, only task1_4.asm program meets the requirements.

• Assembly source code
•  1 assume cs:code
2 code segment
3 start:
4     mov ax, 0b800h
5         mov ds, ax
6         mov bx,0f00h
7         mov cx,50h
8
9 s:    mov [bx],0403h
11     loop s
12
13     mov ah, 4ch
14         int 21h
15 code ends
16 end start
• Screenshot of operation results
• -f b800:0f00 0f9f 03 04 operation results

• Screenshot of task2 running 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,000ah
20
21 s:    mov ax,[bx]
23     mov [bx+20h],ax
24     inc bx
25     loop s
26
27     mov ah,4ch
28     int 21h
29 code ends
30 end start
• Load, disassemble and debug screenshots in debug
• Disassembly:
• Before adding data items in turn, check the debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3

• After adding in sequence, view 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 stack segment
12     dw 8 dup(?)
13 stack ends
14
15 code segment
16 start:
17     mov ax,data1
18     mov ds,ax
19     mov bx,0
20     mov sp,10h
21     mov cx,8
22
23 s0:    push [bx]
25     loop s0
26
27     mov cx,8
28     mov bx,0
29 s1:    pop [bx+10h]
31     loop s1
32
33     mov ah, 4ch
34     int 21h
35 code ends
36 end start
• Load, disassemble and debug screenshots in debug
• Disassembly
•   Before the program is executed, the d command checks the screenshot of the memory space corresponding to the data segment data2
• After the program is executed, the d command views a screenshot of the memory space corresponding to the 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 the screenshot before the program returns (i.e. after ine25 and before line27)
• db 2,3,4,5,6
• db 5 dup(2)

• db 5 dup(5)
• What is the function of line19 in the source code?
• Answer: 0dfh binary is 0000 1101 1111. And operation to change lowercase letters into uppercase letters, and uppercase letters remain uppercase letters.
• What is the purpose of the byte data in the data segment line4 in the source code?
• A: color code, letters are displayed in corresponding colors.

•  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     dw 0
9 data ends
10
11 code segment
12 start:
13     mov ax,data
14     mov ds,ax
15     mov bx,0
16     mov cx,4
17
18 s0:    mov ds:[40h],cx
19     mov si,0
20     mov cx,4
21
22 s:    mov al,[bx+si]
23     or al,00100000b
24     mov [bx+si],al
25     inc si
26     loop s
27
29     mov cx,ds:[40h]
30     loop s0
31
32     mov ah, 4ch
33     int 21h
34 code ends
35 end start
• Load, disassemble and debug screenshots in debug
• Disassembly

• Before executing the program, use the d command to check the memory space corresponding to the data segment
• After the program is executed, use the d command to view the memory space corresponding to the data segment

•  1 assume cs:code, ds:data, es:table
2
3 data segment
4     db '1975', '1976', '1977', '1978', '1979'
5     dd  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             mov bx,0
20             mov si,0
21             mov di,0
22             mov cx,5
23 s0:    mov ax,[bx+si]
24     mov es:[di],ax
26     mov ax,[bx+si]
27     mov es:[di+2h],ax
30     loop s0
31
32     mov di,0
33     mov si,0
34     mov bx,20
35     mov cx,5
36 s1:    mov ax,[bx+si]
37     mov es:[di+5h],ax
39     mov ax,[bx+si]
40     mov es:[di+7h],ax
43     loop s1
44
45     mov di,0
46     mov si,0
47     mov bx,40
48     mov cx,5
49 s2:    mov ax,[bx+si]
50     mov es:[di+10],ax
53     loop s2
54
55     mov cx,5
56     mov di,0
57 s3:    mov ax,es:[di+5]
58     mov dx,es:[di+7]
59     div word ptr es:[di+10]
60     mov es:[di+13],ax
62     loop s3
63
64     mov ah, 4ch
65     int 21h
66 code ends
67 end start
• Debug screenshot
• Screenshot of data segment data information

• View screenshot of original data information of 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

# 2, Experimental summary

•   The actual memory space allocated to the segment by db N dup(0) in the segment is     N/16 is rounded up and multiplied by 16.
•   end start in addition to notifying the compiler of the end of the program, it can also notify the compiler of where the entry of the program is. If the end is not used to indicate the program start label, the program runs from DS+10h at the beginning of the program by default.
•   If the divisor is 8 bits, the divisor is 16 bits, which is stored in AX by default. If the divisor is 16 bits, the divisor is 32 bits, which is stored in DX and AX. DX stores the high 16 bits and AX stores the low 16 bits.
•   The stack can be used to store in reverse order.
• mov ax,[bx+bp] is wrong. bp memory can only be mixed with si or di or used alone.
• In case of multiple cycles, enter the inner cycle, save the cycle times of the outer cycle, and return the outer cycle times to the CX memory when the inner cycle ends.

Posted by casty on Sat, 06 Nov 2021 05:54:02 -0700