I. Experimental Purpose

_Master assembly programming specifications, familiar with the programming environment.

II. EXPERIMENTAL CONTENTS

Write subroutines to convert lower-case letters into upper-case letters in strings (see Textbook Experiment 11).
_2. Write No. 0 interrupt processing program to display "divide error!" in the middle of the screen when division overflow occurs (see Textbook Experiment 12). (Please preview Chapter 12 and complete the experiment. If you don't have enough time, do it after class.)

3. Algorithms and key codes of experimental steps

1. Write subroutines to convert lower-case letters into upper-case letters in strings (see Textbook Experiment 11).

1.1 Algorithmic Steps
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1.2 Key Codes

assume cs:code
data segment
	db "Life is a chain of moments of enjoyment, not only about survival.",0
data ends

code segment
start:mov ax,data
	      mov ds,ax
	      mov si,0     ;Set the starting address to the data address
	      
	      call letterc ;Call a subroutine that converts lowercase letters into uppercase letters in a string
	      
	mov ah,4ch
	int 21h
	
	  letterc:mov al,ds:[si]      ;Remove the address of the data segment
	        cmp al,0          ;Judge whether it is 0 or not
	        je finish          ;If equal to 0, jump to finish Segment, End Continue Conversion
	        and al,11011111B  ;Convert letters to uppercase
	        mov ds:[si],al     ;Return uppercase letters to strings
	        inc si            ;Read the next letter
	        loop letterc      ;Continue subprogram
	 finish:ret           ;The subroutine ends and goes back to the main program
	
code ends
end start

1.3 Screenshots of experimental results

2. Write No. 0 interrupt processing program to display "divide error!" in the middle of the screen when division overflow occurs (see Textbook Experiment 12).

2.1 Algorithmic Steps
Firstly, pseudocode and program framework are written according to program requirements, that is, interrupt processing program is installed first, then vector table is modified, and interrupt processing program is written well.
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_3. Modify the interrupt vector table so that the processing address of interrupt No. 0 points to 0:200.
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2.2 Key Codes

assume cs:code
code segment
	  start:;do0 Installation program, will do0 Copy the code to memory 0:200 place
	      mov ax,0
	      mov es,ax
	      mov di,200h        ;Set up es:di Point to destination address
	      
	      mov ax,cs
	      mov ds,ax
	      mov si,offset do0  ;Set up ds:si Point to the source address
	      
	      mov cx,offset do0end - offset do0     ;Set up cx For transmission length
	      cld         ;Set the transmission direction to be positive
	      rep movsb   ;Automatic Continuous Copy Code
	      
	      ;Setting Interrupt Vector Table,take do0 The entry address is 0.:200 Save it to Table Number 0
	      mov word ptr es:[0*4],200h
	      mov word ptr es:[0*4+2],0
	      
	      ;Division Overflow Segment
	      mov ax,1000h
	      mov bl,1h
	      div bl
	      
	mov ah,4ch
	int 21h
	
	;Display string“ divide error!"In the middle of the screen
	do0: jmp short do0start
	     db 'divide error!'
do0start:mov ax,cs
         mov ds,ax
         mov si,202h          ;Set up ds:si Point to a string
         
         mov ax,0B800h
	     mov es,ax
	     mov di,12*160+36*2   ;Set up es:di Point to the middle of the display memory space
	
	     mov cx,13      ;Set up cx For string length
       s:mov al,[si]
	     mov es:[di],al
	     inc si
	     add di,2
	     loop s
	    
	     mov ax,4c00h
	     int 21h
  do0end:nop
 
code ends
end start	

2.3 Screenshots of experimental results

4. Summary of the experiment (brief description of the problems and solutions in the process of the experiment)

_4.1 The first sub-program is mainly unfamiliar with the program framework. Read more about the book experiment 11 and the related contents of the and and or operation. The actual operation can be done by converting lower-case letters to upper-case letters using and directly converting the corresponding bits.
_4.2 The main difficulty of the second problem lies in the design of program ideas, which can be solved in three steps. The first difficulty is that the length of the program is unknown when the user-defined interrupt handler is copied to the specified memory. The second difficulty is that the strings needed by the user-defined interrupt handler are placed in the program. It can be solved by understanding more about the knowledge mentioned in Chapter 12 of the book and PPT of the teacher in class.