empirical conclusion
Experimental task 1
- Task 1-1
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task1_1.asm source code
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task1_1 screenshot before the end of line17 and line19
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Question answer
- In debug, execute until the end of line17 and before line19. Record this time: register (DS) = 076A, register (SS)= 076B, register (CS) = 076C
- 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-1h_, The segment address of stack is_ X-2h_. (db byte type)
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- Task 1-2
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Task task1_2.asm source code
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task1_2. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values
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Question answer
- In debug, execute until the end of line17 and before line19. Record this time: register (DS) = 076A, register (SS)= 076B, register (CS) = 076C
- 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_.
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- Task 1-3
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Task task1_3.asm source code
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task1_3. Screenshot of the values of registers DS, CS and SS before the end of debugging to line17 and line19
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Question answer
- In debug, execute until the end of line17 and before line19. Record this time: register (DS) = 076A, register (SS)= 076C, register (CS) = 076E
- 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_.
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- Tasks 1-4
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Task task1_4.asm source code
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task1_4. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values
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Question answer
- In debug, execute until the end of line9 and before line11. Record this time: register (DS) = 076C, register (SS) =076E, register (CS) = 076A
- 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+4H_
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- Tasks 1-5
- Based on the practice and observation of the above four experimental tasks, summarize and answer:
- For the segment defined below, after the program is loaded, the actual memory space allocated to the segment is an integer multiple of 16 (if the actual size is N and the loading size is m (M=n*16), then: M-16 < = N < = m).
- 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.
Only program 4 can execute correctly. Because start only identifies the address, and end start tells the compiler that start is the entry. If the end start in the last line is changed to end, the compiler will ignore start and write the default entry address into the executable description information. The default entry address is the first address (low address) of the three segment addresses codeseg+dataseg+stackseg. If the code segment (code) is already in the first segment when we encode, the compiler writes the first segment address (cs) into the executable description information without end start. However, if the stack/data segment is the first segment, the address where the executable description information is written is (ss)/(ds). If the data is parsed into instructions, the program will certainly not run normally, so only the code segment in front can run correctly
Experimental task 2
- Assembly source code
assume cs:code code segment start: mov ax, 0b800h mov ds, ax mov bx, 0f00h mov cx, 50h s: mov ds:[bx], 0403h add bx, 2 loop s mov ah, 4ch int 21h code ends end start
- Screenshot of operation results
Experimental task 3
- Complete assembly source code
assume cs:code data1 segment db 50, 48, 50, 50, 0, 48, 49, 0, 48, 49 ; ten numbers data1 ends data2 segment db 0, 0, 0, 0, 47, 0, 0, 47, 0, 0 ; ten numbers data2 ends data3 segment db 16 dup(0) data3 ends code segment start: mov ax, data1 mov ds, ax mov bx, 0 mov cx, 0ah s: mov ax, ds:[bx] add ax, ds:[bx+10h] mov ds:[bx+20h], ax inc bx loops mov ah, 4ch int 21h code ends end start
- Load, disassemble and debug screenshots in debug
Operation results:
Disassembly:
It is required to give the debug command and screenshot to view the original value of memory space data corresponding to logical segments data1, data2 and data3 before adding data items in turn
And, after adding in turn, view the debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3
Experimental task 4
- Complete assembly source code
assume cs:code data1 segment dw 2, 0, 4, 9, 2, 0, 1, 9 data1 ends data2 segment dw 8 dup(?) data2 ends code segment start: mov ax, data1 mov ds, ax mov sp, 9 mov bx, 0 mov cx, 8 s1: push ds:[bx] add bx, 2 loop s1 mov ax, data2 mov ds, ax mov bx, 0 mov cx, 8 s2: pop ds:[bx] add bx, 2 loop s2 mov ah, 4ch int 21h code ends end start
- Load, disassemble and debug screenshots in debug
It is required to give a screenshot of the memory space corresponding to data segment data2 by using the d command before the program exits.
result:
Experimental task 5
- task5.asm source code
assume cs:code, ds:data data segment db 'Nuist' db 2, 3, 4, 5, 6 data ends code segment start: mov ax, data mov ds, ax mov ax, 0b800H mov es, ax mov cx, 5 mov si, 0 mov di, 0f00h s: mov al, [si] and al, 0dfh mov es:[di], al mov al, [5+si] mov es:[di+1], al inc si add di, 2 loop s mov ah, 4ch int 21h code ends 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)
- What is the function of line19 in the source code?
Convert lowercase letters to uppercase letters
- What is the purpose of the byte data in the data segment line4 in the source code?
Try to modify the value of line4 and find that it stores codes of different colors
Experimental task 6
- task6.asm source code
assume cs:code, ds:data data segment db 'Pink Floyd ' db 'JOAN Baez ' db 'NEIL Young ' db 'Joan Lennon ' data ends code segment start: mov ax, data mov ds, ax mov bx, 0 mov cx, 4 mov si, 0 s1: mov di, cx mov cx, 4 s2: mov al, [bx+si] or al, 20h mov [bx+si], al inc si loop s2 add bx, 10h mov si, 0 mov cx, di loop s1 mov ah, 4ch int 21h code ends end start
- Load, disassemble and debug screenshots in debug
Disassembly
- It is required to give a screenshot of the memory space corresponding to the data segment data by using the d command before the program exits
Experimental task 7
- task7.asm source code
assume cs:code, ds:data, es:table data segment db '1975', '1976', '1977', '1978', '1979' dw 16, 22, 382, 1356, 2390 dw 3, 7, 9, 13, 28 data ends table segment db 5 dup( 16 dup(' ') ) ; table ends code segment start: mov ax, data mov ds, ax mov ax, table mov es, ax mov cx, 5 mov bx, 0 mov di, 0 s: mov al, ds:[bx] mov es:[di], al mov al, ds:[bx + 1] mov es:[di + 1], al mov al, ds:[bx + 2] mov es:[di + 2], al mov al, ds:[bx + 3] mov es:[di + 3], al add bx, 4 add di, 16 loop s mov cx, 5 mov di, 5 s1: mov al, ds:[bx] mov es:[di], al mov ah, ds:[bx + 1] mov es:[di + 1], ah add bx, 2 add di, 16 loop s1 mov cx, 5 mov di, 7 s2: mov al, 0 mov es:[di], al mov es:[di+1], al add di, 16 loop s2 mov cx, 5 mov di, 10 s3: mov al, ds:[bx] mov es:[di], al mov ah, ds:[bx+1] mov es:[di+1], ah add bx, 2 add di, 16 loop s3 mov cx, 5 mov di, 13 mov si, 5 s4: mov al, es:[si] mov ah, es:[si+1] mov dl, es:[si+2] mov dh, es:[si+3] div word ptr es:[si+5] mov es:[di], ax add di, 16 add si, 16 loop s4 mov ah, 4ch int 21h code ends end start
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View screenshot of original data information of table segment
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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