4, Experimental conclusion
1. Experimental task 1
task1.asm source code:
1 assume cs:code, ds:data 2 3 data segment 4 x db 1, 9, 3 5 len1 equ $ - x ; symbolic constants , $Refers to the offset address of the next data item. In this example, it is 3 6 7 y dw 1, 9, 3 8 len2 equ $ - y ; symbolic constants , $Refers to the offset address of the next data item. In this example, it is 9 9 data ends 10 11 code segment 12 start: 13 mov ax, data 14 mov ds, ax 15 16 mov si, offset x ; Take symbol x Corresponding offset address 0 -> si 17 mov cx, len1 ; From symbol x Number of consecutive byte data items at the beginning -> cx 18 mov ah, 2 19 s1:mov dl, [si] 20 or dl, 30h 21 int 21h 22 23 mov dl, ' ' 24 int 21h ; Output space 25 26 inc si 27 loop s1 28 29 mov ah, 2 30 mov dl, 0ah 31 int 21h ; Line feed 32 33 mov si, offset y ; Take symbol y Corresponding offset address 3 -> si 34 mov cx, len2/2 ; From symbol y Number of consecutive word data items started -> cx 35 mov ah, 2 36 s2:mov dx, [si] 37 or dl, 30h 38 int 21h 39 40 mov dl, ' ' 41 int 21h ; Output space 42 43 add si, 2 44 loop s2 45 46 mov ah, 4ch 47 int 21h 48 code ends 49 end start
Operation results:
question answering
Question 1: line27, Assembly instruction loop s1 When jumping, it jumps according to the displacement. Check the machine code through debug disassembly and analyze the jump displacement? (the displacement value is answered in decimal) from the perspective of the CPU, explain how to calculate the offset address of the instruction after the jump label S1.
answer:
debug disassembly result:
The offset of label s1 is 000DH. If the offset of the currently executing instruction plus the space occupied by this instruction is 001BH, the displacement is 001BH - 000DH = 000EH, that is, decimal number 14.
The CPU obtains the displacement by calculating the next instruction address - s1 label address of the LOOP instruction.
Question 2: line44, assembly instructions loop s2 When jumping, it jumps according to the displacement. Check the machine code through debug disassembly and analyze the jump displacement? (the displacement value is answered in decimal) from the perspective of the CPU, explain how to calculate the offset address of the instruction after the jump label S2.
answer:
debug disassembly result:
The offset of label s2 is 0029H. If the offset of the currently executing instruction plus the space occupied by this instruction is 0039H, the displacement is 0039H - 0029H = 0010H, which is the decimal number 16.
The CPU obtains the displacement by calculating the next instruction address - s2 label address of the LOOP instruction.
Question 3: attach the disassembly screenshot of debugging observation in debug during the above analysis.
*Attached to the answer.
2. Experimental task 2
task2.asm source code:
1 assume cs:code, ds:data 2 3 data segment 4 dw 200h, 0h, 230h, 0h 5 data ends 6 7 stack segment 8 db 16 dup(0) 9 stack ends 10 11 code segment 12 start: 13 mov ax, data 14 mov ds, ax 15 16 mov word ptr ds:[0], offset s1 17 mov word ptr ds:[2], offset s2 18 mov ds:[4], cs 19 20 mov ax, stack 21 mov ss, ax 22 mov sp, 16 23 24 call word ptr ds:[0] 25 s1: pop ax 26 27 call dword ptr ds:[2] 28 s2: pop bx 29 pop cx 30 31 mov ah, 4ch 32 int 21h 33 code ends 34 end start
According to the jump principle of call instruction, theoretically, the register ax = 0021h, bx before program execution and exit (Line31)= 0026h，cx = 076ch.
In debug, use the u command to disassemble to get the exit instruction offset address, and use the g command to debug to the exit instruction execution. It is observed that ax = 0021h, bx = 0026h, cx = 076ch.
3. Experimental task 3
task3.asm source code:
assume cs:code, ds:data data segment x db 99, 72, 85, 63, 89, 97, 55 len equ $- x data ends stack segment db 16 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov cx, len mov si, offset x s: mov ah, 0 mov al, ds:[si] call printNumber call printSpace inc si loop s mov ah, 4ch int 21h printNumber: mov bl, 10 div bl mov dl, al mov dh, ah or dl, 30H ;Numeric to character mov ah, 2 int 21h ;Output high mov dl, dh or dl, 30H ;Numeric to character int 21h ;Output low ret printSpace: mov ah, 2 mov dl, ' ' int 21h ;Output space ret code ends end start
Screenshot of running test:
4. Experimental task 4
task4.asm source code:
assume cs:code, ds:data data segment str db 'try' len equ $ - str data ends stack segment db 16 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov ax, 0b800h mov es, ax mov cx, len mov si, 0 mov bl, 2 ;green mov bh, 0 ;first line mov di, bh ;Line beginning address call printStr mov bh, 24 ;Last line ;The space occupied by each row is 00 A0H mov bl, bh mov bh, 0 mov cx, bx mov ax, 0 findLastRow: add ax, 00A0H loop findLastRow mov di, ax mov cx, len mov si, 0 mov bl, 4 ;red call printStr mov ah, 4ch int 21h printStr: s: mov al, ds:[si] mov ah, bl mov es:[di], ax inc si inc di inc di loop s ret code ends end start
Screenshot of running test:
5. Experimental task 5
task5.asm source code:
assume cs:code, ds:data data segment stu_no db '201983290068' len = $ - stu_no data ends stack segment db 16 dup(0) stack ends code segment start: mov ax, data mov ds, ax mov ax, 0b800h mov es, ax mov bl, 17H ;White characters on blue background mov si, 0 call setBgColor mov bh, 24 ;Last line ;The space occupied by each row is 00 A0H mov bl, bh mov bh, 0 mov cx, bx mov ax, 0 findLastRow: add ax, 00A0H loop findLastRow mov di, ax ;Front white thread length 34 mov al, '-' mov cx, 34 mov bl, 17H ;White characters on blue background call printLine mov cx, len mov si, 0 mov bl, 17H ;White characters on blue background call printStr ;Rear white thread length 34 mov al, '-' mov cx, 34 mov bl, 17H ;White characters on blue background call printLine mov ah, 4ch int 21h printStr: s: mov al, ds:[si] mov ah, bl mov es:[di], ax inc si inc di inc di loop s ret printLine: s1: mov al, '-' mov ah, bl mov es:[di], ax inc di inc di loop s1 ret setBgColor: mov cx, 2000 ;80*25 s2: mov al, ' ' mov ah, bl mov es:[si], ax inc si inc si loop s2 ret code ends end start
Screenshot of running test:
5, Experimental summary
Through this experiment, I am familiar with 80 × 25 color character mode display principle. Each line has 160 bytes and 25 lines per page. If you want to output at the specified position, you need to calculate accordingly.
Numbers cannot be displayed directly, so they need to be converted into characters first. or 30h can be used for fast conversion.