1. Experimental task 1
Task 1-1
task1_1.asm source code
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
task1_1 screenshot before the end of line17 and line19
Question answer
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-2___， The segment address of stack is___ X-1_.
Task 1-2
Task 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
Question answer
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-2__， The segment address of stack is__ X-1__.
Task 1-3
Task 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
Question answer
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-4__， The segment address of stack is__ X-2__.
Tasks 1-4
Task 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. 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, 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__, Then, the segment address of the data segment is__ 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__ 16 ((N/16) + 1) bytes _.
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.
task1_4.asm can still run because end indicates the entry of the source program. If there is no end, the program will execute from the beginning, only task1_4.asm starts with a code snippet
2. Experimental task 2
Assembly source code
assume cs:code code segment start: mov ax, 0b800h mov ds, ax mov bx, 0f00h mov cx, 80 s:mov byte ptr [bx], 03h inc bx mov byte ptr [bx], 04h inc bx loop s mov ah, 4ch int 21h code ends end start
Screenshot of operation results
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 cx, 10 s:mov dx, 0 mov ax, data1 mov ds, ax add dx, [bx] mov ax, data2 mov ds, ax add dx, [bx] mov ax, data3 mov ds, ax mov [bx], dx inc bx loop s mov ah, 4ch int 21h code ends end start
Load, disassemble and debug screenshots in debug
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
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(?) 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:push [bx] add bx, 2 loop s 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.
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?
Converts lowercase letters in a word 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 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 s1:mov dx,cx mov si,0 mov cx,4 s:mov al,[bx+si] or al,00100000b mov [bx+si],al inc si loop s add bx,16 mov cx,dx loop s1 mov ah, 4ch int 21h code ends end start
Load, disassemble and debug screenshots in debug
The d command is required to view a screenshot of the memory space corresponding to the data segment data.
7. 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 bx, 0 mov si, 0 mov cx, 5 s0:mov ax, ds:[bx] mov es:[si], ax mov ax, ds:[bx+2] mov es:[si+2], ax add bx, 4 add si,10h loop s0 mov bx, 20 ;Year 4 X5=20 Bytes mov si, 5 mov cx, 5 s1:mov ax, ds:[bx] mov es:[si], ax mov word ptr es:[si+2], 0 ;The high address part of the income word is set to 0 add bx, 2 add si, 10h loop s1 mov cx, 5 mov si, 10 mov bx, 30 s2:mov ax, ds:[bx] mov es:[si], ax add bx, 2 add si, 10h loop s2 mov si, 0 mov cx, 5 s3:mov ax, es:[si+5] mov dx, es:[si+7] div word ptr es:[si+10] mov es:[si+13], ax add si, 10h loop s3 mov ah, 4ch int 21h code ends end start
Debug screenshot
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
8. Experimental summary
dup(?) DUP means duplicate data definitions, that is, copy operands.? Indicates that the defined variable does not specify an initial value, that is, no new data is stored in the defined cell. (instead, prepare for future use, i.e. keep these units)