Experimental task 1
Give the program task1.asm source code, and run screenshots
1 assume cs:code, ds:data 2 3 data segment 4 x db 1, 9, 3 5 len1 equ $ - x 6 7 y dw 1, 9, 3 8 len2 equ $ - y 9 data ends 10 11 code segment 12 start: 13 mov ax, data 14 mov ds, ax 15 16 mov si, offset x 17 mov cx, len1 18 mov ah, 2 19 s1:mov dl, [si] 20 or dl, 30h 21 int 21h 22 23 mov dl, ' ' 24 int 21h 25 26 inc si 27 loop s1 28 29 mov ah, 2 30 mov dl, 0ah 31 int 21h 32 33 mov si, offset y 34 mov cx, len2/2 35 mov ah, 2 36 s2:mov dx, [si] 37 or dl, 30h 38 int 21h 39 40 mov dl, ' ' 41 int 21h 42 43 add si, 2 44 loop s2 45 46 mov ah, 4ch 47 int 21h 48 code ends 49 end start
Answer question ①
① line27, when the assembly instruction loop s1 jumps, it jumps according to the displacement. Check the machine through debug disassembly
What is the jump displacement of the code? (the displacement value is answered in decimal) from the perspective of CPU, it is explained
How to calculate the offset address of the instruction after the jump label s1.
The machine code is E2F2. The eight bit binary form of F2 is 11110010, its complement is 10001110, and the decimal form is - 14, that is, the bit shift is 14
CPU: first modify the value of ip by reading the length of the instruction, ip=ip + instruction length, then execute loop s1 and jump to s1. The displacement is equal to the offset address at s1 minus the offset address of the current ip pointing address
Answer question ②
② line44. When the assembly instruction loop s2 jumps, it jumps according to the displacement. Check the machine through debug disassembly
What is the jump displacement of the code? (the displacement value is answered in decimal) from the perspective of CPU, it is explained
How to calculate the offset address of the instruction after the jump label s2.
The machine code of line44 loop command is E2F0. The eight bit binary form of F0 is 11110000, the complement is 10010000, and the decimal form is - 16, that is, the bit shift is 16
CPU: first, modify the value of ip by reading the length of the instruction, ip=ip + instruction length, then execute loop s2 and jump to s2. The displacement is equal to the offset address at s2 minus the offset address of the current ip pointing address
Question ③
③ Attach the disassembly screenshot of debugging observation in debug during the above analysis
Experimental task 2
Contents of this part:
The program task2.asm source code is given
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
After analysis, debugging and verification, register (ax) =? (bx) = (cx) =? Attach the screenshot of the debugging result interface. ① according to the jump principle of call instruction, first analyze theoretically, register (ax) before program execution and exit (line31)=
? register (bx) =? Register (cx) =?
call word ptr ds:[0] short transfer, push the offset address (ip) of the next instruction into the stack and transfer it to ds:[0] address, i.e. s1, and then pop ax will stack the content to ax;
call dword ptr ds:[2] is an inter segment transfer. The base address and offset address (cs and ip) of the next instruction are pushed onto the stack and transferred to the address of ds:[2], that is, s2. Thereafter, pop bx will stack ip to BX and pop cx will stack cs to cx
call dword ptr ds:[2] is an inter segment transfer. The base address and offset address (cs and ip) of the next instruction are pushed onto the stack and transferred to the address of ds:[2], that is, s2. Thereafter, pop bx will stack ip to BX and pop cx will stack cs to cx
AX=0021 BX=0026 CX=076C
② Assemble and link the source program to get the executable program task2.exe. Use debug to debug, observe and verify debugging
Whether the results are consistent with the theoretical analysis results.
Experimental task 3
Contents of this part:
The program source code task3.asm is given
1 assume cs:code, ds:data 2 3 data segment 4 x db 99, 72, 85, 63, 89, 97, 55 5 len equ $- x 6 data ends 7 8 code segment 9 start: 10 mov ax, data 11 mov ds, ax 12 13 mov cx, len 14 mov si, 0 15 print: 16 mov al, [si] 17 mov ah, 0 18 call printNumber 19 call printSpace 20 inc si 21 loop print 22 23 mov ah, 4ch 24 int 21h 25 26 printNumber: 27 mov bl, 10 28 div bl 29 mov bx, ax 30 31 mov ah, 2 32 33 mov dl, bl ; Printer 34 or dl, 30h 35 int 21h 36 37 mov dl, bh ; Print remainder 38 or dl, 30h 39 int 21h 40 ret 41 42 printSpace: 43 mov ah, 2 44 mov dl, ' ' 45 int 21h 46 ret 47 48 code ends 49 end start
Screenshot of running test
Experimental task 4
Contents of this part:
The program source code task4.asm is given
1 assume cs:code, ds:data 2 3 data segment 4 str db 'try' 5 len equ $ - str 6 data ends 7 8 code segment 9 start: 10 mov ax, data 11 mov ds, ax 12 13 mov bh, 0 ;Specify the first row 14 mov bl, 2 ;Green characters on black background 15 call printStr 16 17 mov bh, 24 ;Specify last line 18 mov bl, 4 ;Scarlet letter on black background 19 call printStr 20 21 mov ah, 4ch 22 int 21h 23 24 printStr: 25 mov al,0ah ;160 bytes per line 26 mul bh ;Multiply by the line number to get the starting address of the line 27 add ax, 0b800h 28 mov es,ax 29 30 mov cx, len ;String length 31 mov si, offset str 32 mov di, si 33 s: 34 mov al, [si] 35 mov ah, bl ;Pass in color attributes 36 mov es:[di], ax ;Put the characters to be displayed into video memory 37 inc si 38 add di, 2 39 loop s 40 ret 41 code ends 42 end start
Screenshot of running test
Experimental task 5
Contents of this part:
The program source code task5.asm is given
1 assume cs:code, ds:data 2 3 data segment 4 stu_no db '201983290032' 5 len = $ - stu_no 6 data ends 7 8 code segment 9 start: 10 mov ax, data 11 mov ds, ax 12 mov di, 0 13 14 call printStuNum 15 16 mov ah, 4ch 17 int 21h 18 19 printStuNum: 20 mov ax, 0b800h 21 mov es, ax 22 mov si, 1 23 24 mov al, 24 25 mov dl, 80 26 mul dl 27 28 mov cx, ax 29 printBlue: 30 mov al, 17h ;00010111 31 mov es:[si], al ;fill color 32 add si, 2 33 loop printBlue 34 35 sub si, 1 36 mov ax, 80 37 sub ax, len 38 mov dl, 2 39 div dl 40 mov dx, ax ;preservation-Length of 41 42 mov cx, dx 43 call printheng 44 45 mov cx, len 46 printStu: ;Output student number 47 mov al, ds:[di] 48 mov ah, 17h 49 mov word ptr es:[si], ax 50 inc di 51 add si, 2 52 loop printStu 53 54 mov cx, dx 55 call printheng 56 57 ret 58 59 printheng: 60 mov al, '-' 61 mov ah, 17h 62 mov word ptr es:[si], ax 63 add si, 2 64 loop printheng 65 ret 66 67 code ends 68 end start
Screenshot of running test
