assume cs:code
data1 segment
    db 50, 48, 50, 50, 0, 48, 49, 0, 48, 49 ; ten numbers
data1 ends

data2 segment
    db 0, 0, 0, 0, 47, 0, 0, 47, 0, 0       ; ten numbers
data2 ends

data3 segment
    db 16 dup(0)
data3 ends

code segment
start:
    mov ax,data3
    mov ds,ax
    mov ax,data1
    mov es,ax

    mov cx,10
    mov bx,0
s1: mov al,es:[bx]
    add [bx],al
    inc bx
    loop s1

    mov ax,data2
    mov es,ax

    mov cx,10
    mov bx,0
s2: mov al,es:[bx]
    add [bx],al
    inc bx
    loop s2

    mov ah,41h
    int 21h

code ends
end start

Screenshot of loading, disassembling and debugging in debug:

  After the data items are added in turn, the debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3:

Experiment task 4:

Complete source code

assume cs:code

data1 segment
    dw 2, 0, 4, 9, 2, 0, 1, 9
data1 ends 

data2 segment
    dw 8 dup(?)
data2 ends

code segment
start:
    
    mov ax,data1
    mov ds,ax

    mov ax,data2
    mov ss,ax
    mov sp,16

    mov cx,8
    mov bx,0

    s:
      push [bx]
      add bx,2
    loop s

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

  Before and after execution

Experimental task 5

Source code:

assume cs:code, ds:data
data segment
        db 'Nuist'
        db 5 dup(5)
data ends

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

        mov ax, 0b800H
        mov es, ax

        mov cx, 5
        mov si, 0
        mov di, 0f00h
s:      mov al, [si]
        and al, 0dfh
        mov es:[di], al
        mov al, [5+si]
        mov es:[di+1], al
        inc si
        add di, 2
        loop s

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

Screenshot of operation results:

   Use the debug tool to debug the program, and use the g command to execute it once before the program returns (i.e. after ine25 and before line27):

What is the function of line19 in the source code?

  A: if the lower address is a lowercase letter, turn it into an uppercase letter. By bitwise and, turn the fifth bit of the binary Ascii code corresponding to the lowercase letter into 0, that is, subtract 32 from the Ascii value to become an uppercase letter.

What is the purpose of the byte data in the data segment line4 in the source code?

  A: set the color of the word.

Experimental task 6

assume cs:code, ds:data

data segment
    db 'Pink Floyd      '
    db 'JOAN Baez       '
    db 'NEIL Young      '
    db 'Joan Lennon     '
data ends

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

    mov cx,4
    mov bx,0
s:  mov dx,[bx]
    mov ax,dx
    add al,32
    mov [bx],ax
    add bx,16
    loop s

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

Load, disassemble and debug screenshots in debug

Experimental task 7

assume cs:code, ds:data, es:table

data segment
    db '1975', '1976', '1977', '1978', '1979' ;20 bytes per line
    dw  16, 22, 382, 1356, 2390
    dw  3, 7, 9, 13, 28 
data ends

table segment
    db 5 dup( 16 dup(' ') )  ;
table ends

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

    mov bx,0
    mov si,0
    mov bp,0
    mov di,0
    

    mov cx,5
s:  mov ax,[bx+si]              ;1975
    mov es:[bp+di],ax    
    add si,2
    add di,2
    mov ax,[bx+si]    
    mov es:[bp+di],ax 

    mov ax,0                    ;[sp]0016[sp]
    add ax,cx
    add ax,cx
    add ax,8
    add si,ax                   ;Here to implement y=8+2n To track this specific si How much,among n by cx Value of
    add di,2
    mov ax,[bx+si]              ;ax Store 16 bit divisor
    mov dl,32
    mov dh,00h
    mov es:[bp+di],dx
    add di,2
    mov dl,00h
    mov dh,al
    mov es:[bp+di],dx
    add di,2
    mov dl,ah
    mov dh,32
    mov es:[bp+di],dx

    add si,10
    add di,2
    mov dx,[bx+si]              ;dl Stored divisor
    mov es:[bp+di],dx

    add di,2
    div dl                      ;al Depositors, ah Deposit remainder
    mov dl,32
    mov dh,al
    mov es:[bp+di],dx
    add di,2
    mov dl,00h
    mov dh,32
    mov es:[bp+di],dx

    add bx,4                    ;Next cycle initialization
    add bp,16
    mov si,0
    mov di,0

    loop s

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

Debug screenshot

(1) View screenshot of original data information of table segment

(2) Run in debug until the program exits, use the d command to view the screenshot of the memory space corresponding to the table segment, and confirm whether the information is structurally written to the specified memory as required