1. Experimental task 1
- Task 1-1
To program task1_1.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results.
assume ds:data, cs:code, ss:stack data segment db 16 dup(0) data ends stack segment db 16 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 16 mov ah, 4ch int 21h code ends end start
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
To program task1_2.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results.
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
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
To program task1_3.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results.
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
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
To program task1_4.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results.
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
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_ (N/16) rounding * 16_.
23 ② if the program Task1_ 1.asm, task1_ 2.asm, task1_ 3.asm, task1_ 4. In ASM, the pseudo instruction end start is changed to
end, which program can still execute correctly? The reasons are analyzed and explained in combination with the conclusions obtained from practical observation.
Only task1_4 can be executed.
Because start and end start appear together, otherwise the program will be executed from the beginning, and only Task1_ The first line is the code snippet.
2. Experimental task 2
Write an assembly source program to realize 160 consecutive bytes to memory units b800:0f00 ~ b800:0f9f, and fill hexadecimal numbers repeatedly in turn
According to 03 04.
assume cs:code code segment start: mov ax,0b800h mov ds,ax mov bx,0f00h mov cx,80 s: mov [bx],0403h add bx,2 loop s mov ax,4c00h int 21h code ends end start
3. Experimental task 3
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 cx,16 mov bx,0 s: mov al,ds:[bx] add al,ds:[bx+16] mov ds:[bx+32],al inc bx loop s mov ah,4ch int 21h code ends end start
The first three data memory spaces:
After operation:
4. Experimental task 4
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 bx,0 mov si,30 mov cx,8 s: mov ax,[bx] mov [si],ax sub si,2 add bx,2 loop s mov ah, 4ch int 21h code ends end start
5. Experimental task 5
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
- Assemble and link the program to get the executable file, run and observe the results.
- Use the debug tool to debug the program and observe the results before the program returns, that is, after line25 and before line27.
- What is the function of line19 in the source code?
Convert letters to uppercase letters
- What is the purpose of the byte data in the data segment line4 in the source code?
Sets the color of the output character
6. Experimental task 6
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: or byte ptr [bx],32 add bx,16 loop s mov ah, 4ch int 21h code ends end start
7. Experimental task 7
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 bx,0 mov bp,0 mov si,20 mov cx,5 s: mov ax,ds:[bx] mov es:[bp],ax mov ax,ds:[bx+2] mov es:[bp+2],ax mov ax,ds:[si] mov es:[bp+5],ax mov word ptr es:[bp+7],0 mov ax,ds:[si+10] mov es:[bp+10],ax mov ax,ds:[si] mov dl,ds:[si+10] div dl mov es:[bp+13],al mov byte ptr es:[bp+14],0 add bx,4 add bp,16 add si,2 loop s mov ah, 4ch int 21h code ends end start
