1. Experimental task 1
1_1)task1_1.asm source code:
assume ds:data, cs:code, ss:stack data segment db 16 dup(0) ;16 byte units are reserved, and the initial values are 0 data ends stack segment db 16 dup(0) ;16 byte units are reserved, and the initial values are 0 stack ends code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 16 ;Set stack top mov ah, 4ch int 21h code ends end start
task1_1 screenshot before the end of line17 and line19
Compile and connect the source program
debug use the d command to check that the memory unit pointed to by the current CS:IP is 076C:0000, and then use the u command to disassemble the corresponding memory to get the address of the target statement as 076C:000A
Use the g command to execute after line17 and before 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-2 and the segment address of the stack is X-1
1_2)task1_2.asm source code
assume ds:data, cs:code, ss:stack data segment db 4 dup(0) data ends stack segment db 8 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 8 mov ah, 4ch int 21h code ends end start
task1_2. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values
The process is consistent with task 1
You can see that the target command address is 076C:000A
Use the g command to execute after line17 and before 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-2 and the segment address of the stack is X-1
1_3)task1_3.asm source code:
assume ds:data, cs:code, ss:stack data segment db 20 dup(0) data ends stack segment db 20 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 20 mov ah, 4ch int 21h code ends end start
task1_3. Screenshot of the values of registers DS, CS and SS before the end of debugging to line17 and line19
Use the g command to execute after line17 and before 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-4 and the segment address of the stack is X-2
1_4)task1_4.asm source code:
assume ds:data, cs:code, ss:stack code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 20 mov ah, 4ch int 21h code ends data segment db 20 dup(0) data ends stack segment db 20 dup(0) stack ends end start
task1_4. Screenshot of the values of registers DS, CS and SS before debugging to line9 and line11
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+2 and the segment address of the stack is X+4
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 ([N/16]+1)*16
xxx segment db N dup(0) 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 still be executed correctly. end start informs the compiler that the entry of the program is at start. Removing start may lead to positioning errors.
With Task1_ 1. Take ASM as an example:
At this time, the CS:IP actually points to the data segment. The u command disassembly finds that the value at CS:IP is the 0 reserved for the data segment, and the program executes it as an instruction, which is obviously wrong
2. Experimental task 2
Assembly source code:
assume cs:code code segment mov ax, 0b800h mov ds, ax mov bx, 0f00h mov cx, 50h ;Cycle 80 times mov ax, 0403h s: mov ds:[bx], ax add bx, 2 loop s mov ah, 4ch int 21h code ends end
Screenshot of operation results
Compile, connect and run the source code
3. 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 bx 0 mov ax data1 mov ds ax mov cx 0ah s: mov ax, [bx] add ax, [bx+10h] ;data1 and data2 Data addition of mov [bx+20h], ax ;put to data3 In paragraph inc bx loop s mov ah, 4ch int 21h code ends end start
Load, disassemble and debug screenshots in debug
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
You can find the perfect addition and save it
4. 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(0) data2 ends code segment start: mov ax,data1 mov ds,ax mov ax,data2 mov ss,ax mov sp,16 mov bx,0 mov cx,8 s: mov ax,[bx] push ax add bx,2 loop s mov ah, 4ch int 21h code ends end start
Load, disassemble and debug screenshots in debug
Before the program exits, use the d command to view a screenshot of the memory space corresponding to the data segment data2
5. 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?
A: the function is to change lowercase letters into uppercase letters
What is the purpose of the byte data in the data segment line4 in the source code?
A: determine the color of the output letters
6. 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 s: mov al,ds:[bx] or al,20h mov ds:[bx],al add bx,16 loop s mov ah, 4ch int 21h code ends end start
Load, disassemble and debug screenshots in debug
Before the program exits, use the d command to view a screenshot of the memory space corresponding to the data segment data
Run until the program exits
Then view the memory space corresponding to the data segment data
You can see that the first uppercase letter of each line is replaced by lowercase
7. Experimental task 7
task7.asm source code
assume cs:code, ds:data, es:table data segment db '1975', '1976', '1977', '1978', '1979' dd 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 bx,0 mov bp,0 mov cx,5 s: mov ax,ds:[bx] mov dx,ds:[bx+2] mov es:[bp],ax mov es:[bp+2],dx mov byte ptr es:[bp+4],' ' add bx,4 add bp,10h loop s mov bp,0 mov cx,5 s1: mov ax,ds:[bx] mov dx,ds:[bx+2] mov es:[bp+5],ax mov es:[bp+7],dx mov byte ptr es:[bp+9],' ' add bx,4 add bp,10h loop s1 mov bp,0 mov cx,5 s2: mov ax,ds:[bx] mov es:[bp+0ah],ax mov byte ptr es:[bp+0ch],' ' add bx,2 add bp,10h loop s2 mov bp,0 mov cx,5 s3: mov ax,es:[bp+5] mov dx,es:[bp+7] div word ptr es:[bp+0ah] mov es:[bp+0dh],ax mov byte ptr es:[bp+0fh],' ' add bp,10h loop s3 mov ah, 4ch int 21h code ends end start
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
You can see the successful write