1. Experimental task 1

(1) Give the program task1.asm source code, and run screenshots  

assume cs:code, ds:data

data segment
    x db 1, 9, 3
    len1 equ $ - x

    y dw 1, 9, 3
    len2 equ $ - y
data ends

code segment
start:
    mov ax, data
    mov ds, ax

    mov si, offset x
    mov cx, len1
    mov ah, 2
 s1:mov dl, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h

    inc si
    loop s1

    mov ah, 2
    mov dl, 0ah
    int 21h

    mov si, offset y
    mov cx, len2/2
    mov ah, 2
 s2:mov dx, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h

    add si, 2
    loop s2

    mov ah, 4ch
    int 21h
code ends
end start

 (2)

Answer question ①   

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.

The displacement is 12   The function of loop label is to cycle jmp short when cx is not less than 0. It is based on the last byte F2 of this instruction. F2 is the complement of - 14, which is the displacement transferred

Answer question ②
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.

The displacement is 14    The function of loop label is to cycle jmp short when cx is not less than 0. It is based on the last byte F0 of this instruction. F0 is the complement of - 16, which is the displacement of transfer

Question ③

③ Attach the disassembly screenshot of debugging observation in debug during the above analysis

 

 

  2. Experimental task 2

(1) The program task2.asm source code is given  

assume cs:code, ds:data

data segment
    dw 200h, 0h, 230h, 0h
data ends

stack segment
    db 16 dup(0)
stack ends

code segment
start:  
    mov ax, data
    mov ds, ax

    mov word ptr ds:[0], offset s1
    mov word ptr ds:[2], offset s2
    mov ds:[4], cs

    mov ax, stack
    mov ss, ax
    mov sp, 16

    call word ptr ds:[0]
s1: pop ax

    call dword ptr ds:[2]
s2: pop bx
    pop cx

    mov ah, 4ch
    int 21h
code ends
end start

(2)

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, it is analyzed theoretically that before the program executes to exit (line31), register (ax) = 0021h, register (bx) = 0026h, register (cx) = 076ch

 

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.

Theoretical analysis: ds:[0] put the offset address of s1 into the stack, and then put it out of the stack into ax. ds:[2] put the offset address of s2 into the stack, and then press the segment address of s2 into the stack, and then put the segment address out into bx, and the offset address into cx. Verify that it is correct.

3. Experimental task 3

(1) The program source code task3.asm is given  

assume cs:code,ds:data
data segment
   x db 99, 72, 85, 63, 89, 97, 55 
   len equ $- x 
data ends

code segment
start:
   mov ax,data
   mov ds,ax
   mov cx,len
   mov si,offset x

s: mov ah,0
   mov al,[si]
   mov bl,10
   div bl
   call printNumber
   call printSpace
   inc si
   loop s

   mov ah,4ch
   int 21h

printNumber:
   mov bx,ax
   mov ah,2
   mov dl,bl
   add dl,30h
   int 21h
   mov dl,bh
   add dl,30h
   int 21h
   ret

printSpace:
   mov ah,2
   mov dl,' '
   int 21h
   ret

code ends
end start

(2) Screenshot of running test

4. Experimental task 4

(1) The program source code task4.asm is given

assume cs:code ds:data

data segment    
  str db 'try'    
  len equ $ - str 
data ends

code segment
start:
  mov ax,data
  mov ds,ax

  mov si,offset str
  mov bh,0
  mov bl,2
  call printStr

  mov si,offset str
  mov bh,24
  mov bl,4
  call printStr

  mov ah,4ch
  int 21h

printStr:
  mov cx,len
  mov ax,0b800h
  mov es,ax
  mov al,0a0h
  mul bh
  mov di,ax

s:mov al,ds:[si]
  mov ah,bl
  mov es:[di],ax
  inc si
  add di,2
  loop s
  ret

code ends
end start

(2) Screenshot of running test

5. Experimental task 5

(1) The program source code task5.asm is given  

 

assume cs:code ds:data

data segment
stu_no db '201983290119' 
len = $ - stu_no 
data ends

code segment
start:
    mov ax,data
    mov ds,ax
    mov bx,0

    mov ax,0b800h
    mov es,ax
    mov cx,80
    mov si,0f00h
    mov dl,'-'
s1: mov es:[si],dl;Fill the last row with all-
    add si,2
    loop s1

    mov cx,len
    mov si,0f44h
s2: mov dl,ds:[bx];Fill the student number from the last line of 68 bytes
    mov es:[si],dl
    add si,2
    inc bx
    loop s2

    mov cx,2000
    mov si,1
    mov dl,17h
s3: mov es:[si],dl;Fill all in blue
    add si,2
    loop s3

    mov ax,4c00h
    int 21h

code ends
end start

(2) Screenshot of running test