Experiment 3 transfer instruction jump principle and its simple application programming

1, Experimental purpose

1. Understand and master the jump principle of transfer instruction

2. Master the method of using call and ret instructions to realize subroutines, and understand and master the parameter transfer mode

3. Understand and master 80 × 25 color character mode display principle

4. Integrate addressing mode and assembly instructions to complete simple application programming

2, Experimental preparation

Review chapters 9-10 of the textbook:

Jump principle of transfer instruction

Usage of assembly instructions jmp, loop, jcxz, call, ret, retf

3, Experimental content

1. Experimental task 1

task1.asm source code:

assume cs:code, ds:data

data segment
    x db 1, 9, 3
    len1 equ $ - x    ; symbolic constants , $Refers to the offset address of the next data item. In this example, it is 3

    y dw 1, 9, 3
    len2 equ $ - y    ; symbolic constants , $Refers to the offset address of the next data item. In this example, it is 9
data ends

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

    mov si, offset x    ; Take symbol x Corresponding offset address 0 -> si
    mov cx, len1    ; From symbol x Number of consecutive byte data items at the beginning -> cx
    mov ah, 2
 s1:mov dl, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h        ; Output space

    inc si
    loop s1

    mov ah, 2
    mov dl, 0ah
    int 21h        ; Line feed

    mov si, offset y    ; Take symbol y Corresponding offset address 3 -> si
    mov cx, len2/2    ; From symbol y Number of consecutive word data items started -> cx
    mov ah, 2
 s2:mov dx, [si]
    or dl, 30h
    int 21h

    mov dl, ' '
    int 21h        ; Output space

    add si, 2
    loop s2

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

Operation screenshot:

Commissioning screenshot:

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.

Jump from 001B to 000D, the displacement is - 14, and the CPU storage content F2 is the complement of - 14

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.

Jump from 0039 to 0029, the displacement is - 16, and the CPU storage content F0 is the complement of - 16

2. Experimental task 2

task2.asm source code:

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

① According to the jump principle of call instruction, it is analyzed theoretically that the register (ax) before the program execution to exit (line31)=

0021 register (bx) = 0026 register (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.

The debugging screenshot shows that the results are consistent.

3. Experimental task 3

task3.asm source code:

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, 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
    div bl
    mov dl, al    
    mov dh, ah    
    or dl, 30h    
    or dh, 30h

    mov ah, 2
    int 21h
    mov dl, dh
    int 21h
    ret

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

code ends
end start

Screenshot of running test:

4. Experimental task 4

task4.asm source code:

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 ax,0b800h
    mov es,ax
    
    mov si,offset printStr
    mov ah,2
    mov bx,0
    call si
    
    mov si,offset printStr
    mov ah,4
    mov bx,0F00H
    call si
    
    mov ah, 4ch
    int 21h

printStr:
    mov cx,len
    mov si,0
s:  
    mov al,[si]
    mov es:[bx+si],ax
    inc si
    add bx,1
    loop s
    ret

code ends
end start

Screenshot of running test:

5. Experimental task 5

task5.asm source code:

assume cs:code,ds:data
data segment
    str db '201983290329'
    len equ $ - str
data ends

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

    mov ax,0b800h
    mov es,ax
    mov cx,0F9Fh
    mov bx,0

s1:    
    mov ah,17h
    mov al,' '
    mov es:[bx],ax
    add bx,2
    loop s1

    mov bx,0F00h
    mov cx,34
s2:    
    mov al,'-'
    mov es:[bx],ax
    add bx,2
    loop s2

    mov cx,len
    mov si,0
s3:    
    mov al,[si]
    inc si
    mov es:[bx],ax
    add bx,2
    loop s3

    mov cx,34

s4:    
    mov al,'-'
    mov es:[bx],ax
    add bx,2
    loop s4

    mov ax,4ch
    int 21h

code ends
end

Screenshot of running test:

4, Experimental summary

Master the method of using instructions to realize subroutines, feel the transfer process of parameters, and be familiar with the usage of assembly instructions such as jmp, loop, jcxz, call, ret, retf, etc

Posted by dietkinnie on Mon, 29 Nov 2021 21:38:46 -0800

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Experiment 3 transfer instruction jump principle and its simple application programming

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