Experiment 3 transfer instruction jump principle and its simple application programming

task1

Using any text editor, enter the 8086 assembler source code task1.asm.
task1.asm

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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 

Assemble and link the source program to get the executable program task1.exe. After running, combined with the running results, comments and necessary debug Commissioning: 1. Understand the flexible use of operator offset, pseudo instruction equ and predefined symbol $. Continuous data items can be easily calculated through line5, line8 and data attributes (bytes, words, doublewords, etc.) of data items Without manual counting. Note *: the symbolic constants len1 and len2 do not occupy the memory space of the data segment Operation results:

1. question answering
   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
   A: the jump displacement is 14byte,
   As shown in the figure: 1B + [(F2) supplement] original = 27-14=0DH

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.
A: jump to 16byte
As shown in the figure: 39 + [(F0) supplement] original = 57-16=29H

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

task3

Using any text editor, enter the 8086 assembler source code task2.asm.
task3.asm

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DATAS SEGMENT
    ;Enter segment code here  
DATAS ENDS
data segment
    x db 99, 72, 85, 63, 89, 97, 55
    len equ $ - x
data ends
STACKS SEGMENT
    ;Enter the stack segment code here
STACKS ENDS

CODES SEGMENT
    ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
        mov ax,data
        mov ds,ax
        mov cx,len    
        mov si,0
         

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  ;Binary divided by ten, the remainder is single digits, and the quotient is ten digits
        div bl     ;ah Is the remainder, al For quotient( ah，al Respectively ax High 8 bit Low 8 bit
        mov bx,ax  ;Transmit data to bx For modification ax use int 21h

        mov ah,2

        mov dl,bl      
        or dl,30h
        int 21h

        mov dl,bh      
        or  dl,30h ;Digital conversion ASCALL code
        int 21h
        ret

printSpace:
          mov ah,2
          mov dl,' '
          int 21h
          ret
CODES ENDS
    END START

Operation screenshot:

task4

Experiment code:

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DATAS SEGMENT
    ;Enter segment code here  
DATAS ENDS
data segment
    x db 'try' 
    len equ $ - str
    
data ends

STACKS SEGMENT
    ;Enter the stack segment code here
STACKS ENDS

CODES SEGMENT
    ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
    mov ax, data
    mov ds, ax
    mov ax,0b800h
    mov es,ax

    mov si,offset x
    mov bl,     2    ;Specifies that the string color is green on a black background
    mov bh,    0    ;Specify the first line of the behavior

    call printStr

    mov si,offset x
    mov bl,   4       ; Specifies that the string color is red on a black background
    mov bh, 24      ;Specify the last line of the behavior

    call printStr

    mov ah, 4ch
    int 21h
    
printStr:
    mov al, 160
    mul bh

    mov cx,len
    mov di, ax
s:
    mov ah, ds:[si]
    mov es:[di], ah
    inc di
    mov es:[di], bl
    inc si
    inc di
    loop s
    ret

CODES ENDS
    END START

Operation results:

task5

Experiment code:

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DATAS SEGMENT
    ;Enter segment code here  
DATAS ENDS
data segment
    stu_no db '2019832900'
    len = $ - stu_no
data ends
STACKS SEGMENT
    ;Enter the stack segment code here
STACKS ENDS

CODES SEGMENT
    ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
      mov ax,data
    mov ds,ax
    mov ax,0b800h
    mov es,ax
    call p1 ;Set background

    mov bh,24 ;Set the number of rows to the last row
    mov al,160
    mul bh
    mov bx,ax
    call p2 ;Output polyline
    call p3	;Output student number
    call p2	;Output discount

    mov ax,4c00h
    int 21h

p2:
    mov al,'-'
    mov dl,17h;'-'Color of
    mov cx,33;Draw 33 on one side'-'
    s:
    mov es:[bx],al
    inc bx
    mov es:[bx],dl
    inc bx
    loop s
ret


p1:
    mov si,1
    mov bl,17h;Set color
    mov cx,7d0h ;2000 in total w，4000 individual byte(160*25)
    s2:
    mov es:[si],bl
    add si,2
    loop s2
ret

p3:
    mov si,0 ;str'Student number'Offset address of
    mov dl,17h;colour
    mov cx,len;length
    s1:
    mov al,ds:[si]
    mov es:[bx],al
    inc bx
    mov es:[bx],dl
    inc bx
    inc si
    loop s1
ret

CODES ENDS
    END START

Experimental results: