Task 1
Input 8086 assembler source code task1.asm
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
① line27, when the assembly instruction loop s1 jumps, 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.
After the cpu reads the loop s1 instruction, the IP value is 001Bh, and it needs to jump to the address of 000Dh. Therefore, the displacement is: target address - current address = - 14
② line44. When the assembly instruction loop s2 jumps, 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.
After the cpu reads the loop s2 instruction, the IP value is 0039h, and it needs to jump to address 0029h, so the displacement is: - 16
③ Attach the disassembly screenshot of debugging observation in debug during the above analysis
As shown above
Task 2
Enter 8086 assembler source code task2.asm
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
① According to the jump principle of call instruction, it is analyzed theoretically that before the program executes to exit (line31), register (ax) =? Register (bx) =? Register (cx) =?
ax = s1，bx = s2，cx = cs
② Assemble and link the source program to get the executable program task2.exe. Use debug to observe and verify whether the debugging results are consistent with the theoretical analysis results.
From the above figure, s1=0021h, s2=0026h, cs=0772h
The results are consistent with the theoretical analysis
Task three
For 8086 CPU, the known logical segment is defined as follows:
1 data segment 2 x db 99, 72, 85, 63, 89, 97, 55 3 len equ $- x 4 data ends
Write 8086 assembly source program task3.asm, and output this group of continuous data in the data section in decimal form on the screen. The data is separated by spaces
1 assume cs:code, ds:data, ss:stack 2 data segment 3 x db 99, 72, 85, 63, 89, 97, 55 4 len equ $- x 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 mov ax,0b800h 16 mov es,ax 17 18 19 mov bx,0 20 mov si,0 21 22 mov dx,offset printNumber 23 call dx ;Print character 24 mov bx,0 25 mov si,4 26 mov dx,offset printSpace 27 call dx ;Print spaces 28 29 mov ah,4ch 30 int 21h 31 32 printNumber: 33 mov cx,7h 34 s1: mov al,ds:[bx] 35 mov ah,0 36 mov dx,10 ;Separate ten bits from one bit 37 div dl 38 mov dl,10h 39 mov dh,ah 40 or al,30h 41 ;Print ten digits 42 mov ah,7 43 mov es:[si],ax 44 inc bx 45 add si,2 46 ;Print everyone 47 mov al,dh 48 or al,30h 49 mov es:[si],ax 50 add si,4 51 52 loop s1 53 ret 54 printSpace: 55 mov cx,7h 56 s2: mov al,' ' 57 mov ah,7 58 mov es:[si],ax 59 add si,6 60 loop s2 61 ret 62 code ends 63 end start
Task 4
For 8086 CPU, the known logical segment is defined as follows:
1 data segment 2 str db 'try' 3 len equ $ - str 4 data ends
Write the 8086 assembly source program task4.asm, specify the color and line on the screen, and output the string on the screen:
1 assume cs:code, ds:data 2 3 data segment 4 s db 'try' 5 len equ $ - s 6 data ends 7 8 code segment 9 start: 10 mov ax,data 11 mov ds,ax 12 mov si,offset s 13 mov bp,0 14 mov bl,2 15 mov bh,0 16 17 mov dx,printStr 18 call dx 19 mov si,offset s 20 mov bp,0 21 mov bl,4 22 mov bh,0f0h 23 mov dx,printStr 24 call dx 25 26 mov ah,4c 27 int 21h 28 29 30 printStr: 31 mov ah,0b8h 32 mov al,bh 33 mov es,ax 34 mov cx,len 35 s1: mov dl,ds:[si] 36 mov es:[bp],dl 37 inc bp 38 mov es:[bp],bl 39 inc bp 40 inc si 41 loop s1 42 ret 43 44 code ends 45 end start
Task 5
For 8086 CPU, for 8086 CPU, the known logical segment is defined as follows
1 data segment 2 stu_no db '201983290497' 3 len = $ - stu_no 4 data ends
At 80 × In 25 color character mode, the student number is displayed in the middle of the last line of the screen. It is required that the student number and broken lines on both sides of the output window are displayed in white foreground color on a blue background:
1 assume cs:code, ds:data 2 data segment 3 stu_no db '201983290497' 4 len = $ - stu_no 5 data ends 6 7 code segment 8 start: 9 mov ax,data 10 mov ds,ax 11 mov ax,0b800h 12 mov es,ax 13 mov si,offset stu_no 14 15 mov cx,780h 16 mov ah,10h 17 mov al,' ' 18 mov bx,0 19 s1: mov es:[bx],ax ;Dye all except the last line with a blue background 20 add bx,2 21 loop s1 22 23 mov dx,offset print 24 call dx 25 mov cx,len 26 mov ah,17h ;The binary of white characters on blue background is 00010111,16 Hexadecimal 27 27 s3: mov al,ds:[si] 28 mov es:[bx],ax 29 add bx,2 30 inc si 31 loop s3 32 33 call dx 34 mov ah,4ch 35 int 21h 36 37 print: 38 mov cx,22h 39 mov ah,17h 40 mov al,'-' 41 s2: mov es:[bx],ax 42 add bx,2 43 loop s2 44 ret 45 46 code ends 47 end start
summary
① Through the comprehensive utilization of call, ret and addressing methods, the function of a certain section of code can be regarded as a function and can be called in the main code
② B8000H~BFFFFH display buffer has a total space of 32KB. 80x25 color mode indicates that 25 characters can be displayed in total and 80 characters per line. After setting b8, the address is calculated from the first character position of the first line to the last character position ff of the last line