Experimental task 1
Task 1-1
task1_1.asm source code task1_1 screenshot before the end of line17 and line19
Question answer
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_ X-2_， The segment address of stack is_ X-1_.
Task 1-2
task1_2.asm source code task1_2. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values
Question answer
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_ X-2_， The segment address of stack is_ X-1_.
Task 1-3
Task task1_3.asm source code task1_3. Screenshot of the values of registers DS, CS and SS before the end of debugging to line17 and line19
Question answer
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_ X-4_， The segment address of stack is_ X-2_.
Tasks 1-4
Task task1_4.asm source code task1_4. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values
Question answer
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_ X+2_， The segment address of stack is_ X+4_.
Tasks 1-5
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_ int((N+1)/16)+1_.
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 can run. The start after end indicates that the starting place of the source program is start. If there is no start, the program runs from the beginning, while the beginning of the other three programs are data segments and cannot run. Only task1_4 starts with a code segment that can run.
Experimental task 2
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,80 8 9 s: 10 mov [bx],0403H 11 add bx,2 12 loop s 13 14 mov ah,4ch 15 int 21h 16 code ends 17 end start
Experimental results:
Experimental task 3
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,10 20 s: 21 mov al,ds:[bx] 22 add al,ds:[bx+16] 23 mov [bx+32],al 24 inc bx 25 loop s 26 27 mov ah,4ch 28 int 21h 29 code ends 30 end start
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:
Experimental task 4
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 bx,0 16 mov ax,data2 17 mov ss,ax 18 mov sp,10h 19 mov cx,8 20 s: 21 push [bx] 22 add bx,2 23 loop s 24 25 mov ah, 4ch 26 int 21h 27 code ends 28 end start
Screenshot of original data segment:
Before the program exits, use the d command to view a screenshot of the memory space corresponding to the data segment data2:
Experimental task 5
Source code:
1 assume cs:code, ds:data 2 data segment 3 db 'Nuist' 4 db 5 dup(2) 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 24 add di, 2 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):
Q: what is the role of line19 in the source code?
A: use the and method to convert all letters to uppercase letters.
Q: what is the purpose of the byte data in the data segment line4 in the source code?
A: make the five letters appear in different colors.
Modify the value of 5 byte units in line4, reassemble, link, run and observe the results.
Experimental task 6
Source code:
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: 17 mov al,[bx] 18 or al,00100000b 19 mov [bx],al 20 add bx,16 21 loop s 22 23 mov ah, 4ch 24 int 21h 25 code ends 26 end start
Screenshot of original data:
Before the program exits, use the d command to view a screenshot of the memory space corresponding to the data segment data:
Experimental task 7
Source code:
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 mov bx,0 20 mov si,0 21 mov cx,5 22 s1: 23 mov ax,[bx] 24 mov es:[si],ax 25 mov ax,[bx+2] 26 mov es:[si+2],ax 27 add bx,4 28 add si,16 29 loop s1 30 31 mov bx,20 32 mov cx,5 33 mov si,5 34 s2: 35 mov ax,0 36 mov es:[si+2],ax 37 mov ax,[bx] 38 mov es:[si],ax 39 add bx,2 40 add si,16 41 loop s2 42 43 mov bx,30 44 mov cx,5 45 mov si,10 46 s3: 47 mov ax,[bx] 48 mov es:[si],ax 49 add bx,2 50 add si,16 51 loop s3 52 53 mov cx,5 54 mov si,0 55 s4: 56 mov ax, es:[si+5] 57 mov dx, es:[si+7] 58 div word ptr es:[si+10] 59 mov es:[si+13],ax 60 add si,16 61 loop s4 62 63 mov ah, 4ch 64 int 21h 65 code ends 66 end start
View the screenshot of the original data information of the table segment:
Run 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:
As shown in the figure, it has been written successfully.
Experimental summary
Through this experiment, I have a deeper understanding of the contents of chapters 5-8 in the compilation book, I am familiar with how to achieve the desired results through the code written by myself, and I am familiar with the programming process of compilation. Experiment 1 makes me understand the storage mode of memory through reading the program, and also provides a lot of inspiration for me to write the code later. Experiment 2-6 is basically the same, They all use their knowledge flexibly and complete the corresponding topics through programming. Experiment 7 is a little difficult, but through careful analysis and understanding, they divide the problems into parts and solve them step by step, and finally get the ideal results. This experiment enabled me to apply the knowledge in books in practice, which benefited me a lot.