Chapter V exercises (124, 678, 15, 19, 22) [microcomputer principles] [exercises]

Preface

The following is derived from the principle of Microcomputer （ The Fourth Edition ） Edited by Wang Zhongmin
For learning and communication only
Please read the article statement , By default, I agree with this statement

exercises

1

1 Draw a diagram to illustrate the storage space allocated by the following statements and the initialized data values .
(1) BYTE_VAR DB 'BYTE’, 12,-12H,3 DUP(0,7,2DUP(1,2),7)
(2) WORD_ VAR DW 5 DUP(0,1,2),7,-5, ‘BY’, ‘TE’, 256H

1.asm

data segment
	BYTE_VAR	DB	'BYTE',12,-12H,3 DUP(0,7,2 DUP(1,2),7)
	org 40h
	WORD_VAR DW 	5 DUP(0,1,2),7,-5,'BY','TE',256H
data ends

code segment
     assume cs:code,ds:data
start:     
     mov ax,data
     mov ds,ax
 
     mov ah,4ch
     int 21h
    
code ends
     end start

2

Suppose the data in the program is defined as follows :

PARTNO DW ?
PNAME DB 16 DUP(?)
COUNT DD ?
PLENTH EQU $-PARTNO;16h=22

be PLENTH What is the value of ? What does it mean ?

PLENTH The value of is 0016h=22, It indicates the number of currently allocated cell space ;
PARTNO  2
PNAME  16
COUNT  4

2.asm

data segment
	PARTNO DW ?
	PNAME DB 16 DUP(?)
	COUNT DD ?
	PLENTH EQU $-PARTNO
data ends

code segment
     assume cs:code,ds:data
start:     
     mov ax,data
     mov ds,ax
  
    mov ax,PLENTH	 	
  	
     mov ah,4ch
     int 21h
    
code ends
     end start

4

4. Suppose the data in the program is defined as follows

LNAME DB 30 DUP(?)
ADDRESS DB 30 DUP(?)
CITY DB 15 DUP(?)
CODE_LIST DB 1,7,8,3,2

(1) Use one MOV Instructions will be LNAME The offset address of is put into BX.
(2) Use an instruction to put CODE_LIST The contents of the first two bytes of are put into SI.
(3) Write a pseudo instruction definer to make CODB_LENGHT The value is equal to the CODE_LIST The actual length of the field .
4.asm

data segment
	LNAME DB 30 DUP(?)
	ADDRESS DB 30 DUP(?)
	CITY DB 15 DUP(?)
	CODE_LIST DB 1,7,8,3,2
	;(3) Write a pseudo instruction definer to make CODB_LENGHT The value is equal to the CODE_LIST The actual length of the field .
	CODB_LENGHT EQU $-CODE_LIST
data ends

code segment
     assume cs:code,ds:data
start:     
     	mov ax,data
    	 mov ds,ax
	; verification (3)
	mov ax,CODB_LENGHT
  	;(1) Use one MOV Instructions will be LNAME The offset address of is put into BX.
	mov bx,offset LNAME
	;(2) Use an instruction to put CODE_LIST The contents of the first two bytes of are put into SI.
	mov si,WORD PTR CODE_LIST
	
  	
    	 mov ah,4ch
    	 int 21h
    
code ends
     end start

6

6. For the following data definitions , Articles MOV After the instruction is executed separately , What is the content of the register ?

 FLDB DB?
TABLEA DW 20DUP(?)
TABLEB DB ‘ABCD'

(1) MOV AX, TYPE FLDB ;(AX)=1
(2) MOV AX, TYPE TABLEA ;(AX)=2
(3) MOV CX, LENGTH TABLEA ;(CX)=20
(4) MOV DX, SIZE TABLEA ;(DX)=40
(5) MOV CX, LENGTH TABLEB ;(CX)=1; Unused DUP

6.asm

data segment
	FLDB DB ?
	TABLEA DW 20 DUP(?)
	TABLEB DB 'ABCD'
data ends

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

	MOV AX, TYPE FLDB 		;(AX)=1
	MOV AX, TYPE TABLEA 		;(AX)=2
	MOV CX, LENGTH TABLEA 	;(CX)=20
	MOV DX, SIZE TABLEA 		;(DX)=40
	MOV CX, LENGTH TABLEB 	;(CX)=1; Unused DUP
	
  	
    	 mov ah,4ch
    	 int 21h
    
code ends
     end start

7

7. Try to explain which of the following instructions need to be added PTR Pseudo instruction definer

BVAL DB 10H,20H
WVAL DW 1000H

(1) MOV AL, BVAL
(2) MOV DL, [BX]
(3) SUB [BX],2
(4) MOV CL,WVAL
7.asm

data segment
	BVAL DB 10H,20H
	WVAL DW 1000H
data ends

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

	MOV AL,BVAL
	MOV DL,[BX]
	SUB [BX],BYTE PTR  2
	MOV CL,BYTE PTR WVAL
	
  	
    	 mov ah,4ch
    	 int 21h
    
code ends
     end start

8

8. Write a macro definition BXCHG, Set the height of one byte 4 Bit and low 4 Bit exchange .

data segment
	NUM1 DB 12H;-->21
	NUM2 DB 34H;-->43
	NUM3 DB 56H;-->65
data ends

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

BXCHG MACRO NUM	
	mov cl,4
	rol NUM,cl; Cyclic shift to the left 4 position 
	ENDM	
		
	BXCHG NUM1
	BXCHG NUM2
  	BXCHG NUM3

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

15

15. Try to write an assembly language program , Request to receive a... From the keyboard 4 Hexadecimal number of bits , And display on the terminal
Show its binary equivalent .

Not 16 Hexadecimal characters press 0 Handle

data segment
	INF1 DB "Please input a hex mumber :$"
	INF2 DB 0AH,0DH,"$"
	INF3 DB " $"
	org 60H
	IBUF DB 7,0,6 DUP('$')
	
data ends

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

; Output INFO
PINF MACRO INF
MOV	DX,OFFSET INF
	MOV	AH,09H
	INT	21H
ENDM
; Take out each bit and output 
QUWEI MACRO NUM
	MOV CX,4
	SHL NUM,CL; To shift four digits left to a significant number ,0f->f0
	LOOP1:
		SHL NUM,1
		JC P1
		JNC P0
		P1:
			PUSH DX;MOV DL,'1' Will change 
			MOV AH,2
			MOV DL,'1'		
			int 21h
			POP DX
			JMP END1
		P0:
			PUSH DX
			MOV AH,2
			MOV DL,'0'		
			int 21h
			POP DX
			JMP END1
		END1:
			NOP
	LOOP LOOP1
ENDM	
; Output BUF
PBUF MACRO BUF
	MOV CX,0
	MOV CL,[BUF+1]
	LEA BX, BUF+2
LOOP3:
	PUSH CX;QUWEI DL; Will change CX
	;MOV DL,[BX]
	;ADD DL,30H
	;MOV AH, 02H
	;INT 21H
	MOV DL,[BX]
	QUWEI DL
	INC BX
	POP CX
	PINF INF3; One bit per revolution 16 Base to 4 position 2 Base to a space 
LOOP LOOP3
ENDM

; Input --》BUF
STORE MACRO BUF
	MOV	DX, OFFSET IBUF	; Type a hexadecimal number 
	MOV	AH,0AH
	INT	21H
ENDM

	PINF  INF1
	CALL  HTB
	PINF  INF2
	PBUF IBUF	

    	 mov ah,4ch
    	 int 21h
  


;DTB rewrite HTB   
;MOV DX,10-->MOV DX,16
HTB  PROC NEAR
	STORE IBUF

	MOV	CL,IBUF+1; The number of decimal digits is sent to CX
	MOV	CH,0
	MOV	SI,OFFSET IBUF+2; Point to the first character of the input ( highest )
	
	MOV	AX,0	; Start converting decimal numbers to binary numbers 
	AGAIN: 
		;MOV	DX,10;((0x10+a4)x10+...)x10+a0
		MOV	DX,16;((0x16+a4)x16+...)x16+a0
		MUL	DX
		;AND	BYTE PTR [SI],0FH; The decimal system can be converted into numbers 
		;16 The base number needs to be changed 
		CMP BYTE PTR [SI],'0'
		JB  OTHER
		CMP  BYTE PTR [SI],'9'
		JBE  DIGIT
		CMP  BYTE PTR [SI],'A'
		JB  OTHER
		CMP  BYTE PTR [SI],'F'
		JBE  UPPER
		CMP  BYTE PTR [SI],'a'
		JB  OTHER
		CMP  BYTE PTR [SI],'f'
		JBE LOWER
		JMP PEND
		LOWER:
			; Lowercase letters a->10
			SUB BYTE PTR [SI],'a'
			ADD BYTE PTR [SI],10
			JMP PEND; Note that the program should jump out after executing a branch 
		UPPER:
			; Capital A->10
			SUB BYTE PTR [SI],'A'
			ADD BYTE PTR [SI],10
			JMP PEND
		DIGIT:  
			AND	BYTE PTR [SI],0FH; The situation of numbers 
			JMP PEND
		OTHER:
			AND	BYTE PTR [SI],0H; Input not 16 Hexadecimal characters press 0 Handle 
			JMP PEND
		PEND:		

			ADD	AL, [SI]
			ADC	AH,0
			INC	SI
	LOOP	AGAIN
	RET
HTB ENDP


code ends
     end start

19

19. Enter a series of characters from the keyboard ( End with carriage return ), And press letters 、 Count numbers and other characters by category , most
The counting results of these three categories are displayed .

19.asm

data segment
	INF1 DB "Please input a string :$"
	INF2 DB 0AH,0DH,"$"
	INF3 DB " $"

	org 40H
	CNTN DB 0  ; Number of numbers 
	CNTC DB 0  ; Number of letters 
	CNTO DB 0 ; Other numbers 

	org 60H
	IBUF DB 255,0,254 DUP('$')
	OBUF DB 7,0,6 DUP('$')
data ends

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

; Output INFO
PINF MACRO INF
MOV	DX,OFFSET INF
	MOV	AH,09H
	INT	21H
ENDM

; Input --》BUF
STORE MACRO BUF
	MOV	DX, OFFSET IBUF	; Type a hexadecimal number 
	MOV	AH,0AH
	INT	21H
ENDM

; Output BUF
PBUF MACRO BUF
	MOV CX,0
	MOV CL,[BUF+1]
	LEA BX, BUF+2
LOOP3:
	
	MOV DL,[BX]
	ADD DL,30H
	MOV AH, 02H
	INT 21H
	INC BX	
LOOP LOOP3
ENDM

; Statistics 
CNT  MACRO
	STORE IBUF

	MOV	CL,IBUF+1; The number of decimal digits is sent to CX
	MOV	CH,0
	MOV	SI,OFFSET IBUF+2; Point to the first character of the input ( highest )
	
	MOV	AX,0	; Start converting decimal numbers to binary numbers 
	AGAIN: 
		
		CMP BYTE PTR [SI],'0'
		JB  OTHER
		CMP  BYTE PTR [SI],'9'
		JBE  DIGIT
		CMP  BYTE PTR [SI],'A'
		JB  OTHER
		CMP  BYTE PTR [SI],'Z' 
		JBE  UPPER
		CMP  BYTE PTR [SI],'a'
		JB  OTHER
		CMP  BYTE PTR [SI],'z'
		JBE LOWER
		CMP  BYTE PTR [SI],'z'
		JA OTHER
		JMP PEND
		LOWER:
			INC CNTC
			JMP PEND; Note that the program should jump out after executing a branch 
		UPPER:			
			INC CNTC
			JMP PEND
		DIGIT:  
			INC CNTN
			JMP PEND
		OTHER:
			INC	CNTO
			JMP PEND
		PEND:		
			ADD	AL, [SI]
			ADC	AH,0
			INC	SI
	LOOP	AGAIN

ENDM

;AL->BCD; compressibility 
HTBCD MACRO
	mov ah,0
	
	mov cl,4
	push ax
	shr al,cl
	
	mov bl,16h
	mul bl
	
	pop bx
	and bl,0fh
	add al,bl
	daa
	
ENDM

; Output BCD code ;AL compressibility 	
PBCD MACRO
	mov ah,0
	mov cl,4
	push ax
	shr al,cl		
	mov bh,al		;al High position 
	pop ax
	and al,0fh		;al Low position 
	mov ah,bh		;al High position 

	push ax
	mov dl,ah
	add dl,30h
	mov ah,2
	int 21h
	pop ax

	push ax
	mov dl,al
	add dl,30h
	mov ah,2
	int 21h
	pop ax
ENDM
	
	PINF  INF1
	CNT

	PINF  INF2
	mov ax,0
	mov al,CNTN
	HTBCD 
	PBCD

	PINF  INF3
	mov al,CNTC
	HTBCD 
	PBCD

	PINF  INF3
	mov al,CNTO
	HTBCD 
	PBCD

    	 mov ah,4ch
    	 int 21h
  



code ends
     end start

22

22. Write two 4 Bit uncompressed BCD Sum of the numbers , The program that sends the and to the display .
22.asm

data segment
	INF1 DB "RESULT:$"
	INF2 DB 0AH,0DH,"$"
	INF3 DB " $"

	org 20H
	STRING1 DW 0001H
	STRING2 DW 0019H
	D DB 0h			;0h
	SUM DW 0H		;0020h

	;STRING1 DW 0001H
	;STRING2 DW 0099H
	;D DB 0h			;0h
	;SUM DW 0H		;0100h

	;STRING1 DW 0001H
	;STRING2 DW 9999H
	;D DB 0h			;1h
	;SUM DW 0H;0020h	;0000h


	org 60H
	IBUF DB 255,0,254 DUP('$')
	OBUF DB 7,0,6 DUP('$')
data ends

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

; Output INFO
PINF MACRO INF
MOV	DX,OFFSET INF
	MOV	AH,09H
	INT	21H
ENDM

; Input --》BUF
STORE MACRO BUF
	MOV	DX, OFFSET IBUF	; Type a hexadecimal number 
	MOV	AH,0AH
	INT	21H
ENDM


; Take out bx:BCD Count each bit and output 
QUWEI MACRO 
	PUSH BX
	MOV CX,12
	SHR BX,CL			
	MOV DL,BL
	ADD DL,30H
	MOV AH,02H
	int 21h
	POP BX

	PUSH BX
	MOV CX,8
	SHR BX,CL	
	AND BX,0FH		
	MOV DL,BL
	ADD DL,30H
	MOV AH,02H
	int 21h
	POP BX

	PUSH BX
	MOV CX,4
	SHR BX,CL	
	AND BX,0FH		
	MOV DL,BL
	ADD DL,30H
	MOV AH,02H
	int 21h
	POP BX

	PUSH BX
	AND BX,0FH		
	MOV DL,BL
	ADD DL,30H
	MOV AH,02H
	int 21h
	POP BX	
	
ENDM	

	PINF INF1
	PINF INF2



	; Calculation 
	;ax,bx On behalf of the digital 
	;cl, Express al+bl Carry of 
	;ch, Express ah+bh Carry of 
	mov ax,STRING1
	mov bx,STRING2
	mov cx,0
	push ax

	add al,bl
	daa
	adc cl,0
	mov dx,ax
	
	
	pop ax
	; The carry of the lower order is required 
	add ah,cl
	add ah,bh
	mov al,ah
	daa
	adc ch,0
	mov dh,al
	; The result is  ch ,dx
	mov D,ch
	mov SUM,dx

	; Show 
	mov dl,D
	add dl,30H
	mov ah,2
	int 21h
	
	mov bx,sum

	QUWEI
	

    	 mov ah,4ch
    	 int 21h
  



code ends
     end start

simplify QUWEI

data segment
	INF1 DB "RESULT:$"
	INF2 DB 0AH,0DH,"$"
	INF3 DB " $"

	org 20H
	STRING1 DW 0001H
	STRING2 DW 0019H
	D DB 0h			;0h
	SUM DW 0H		;0020h

	;STRING1 DW 0001H
	;STRING2 DW 0099H
	;D DB 0h			;0h
	;SUM DW 0H		;0100h

	;STRING1 DW 0001H
	;STRING2 DW 9999H
	;D DB 0h			;1h
	;SUM DW 0H;0020h	;0000h


	org 60H
	IBUF DB 255,0,254 DUP('$')
	OBUF DB 7,0,6 DUP('$')
data ends

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

; Output INFO
PINF MACRO INF
MOV	DX,OFFSET INF
	MOV	AH,09H
	INT	21H
ENDM

; Input --》BUF
STORE MACRO BUF
	MOV	DX, OFFSET IBUF	; Type a hexadecimal number 
	MOV	AH,0AH
	INT	21H
ENDM


PDL MACRO
	ADD DL,30H
	MOV AH,02H
	int 21h

ENDM

; Take out bx:BCD Count each bit and output 
QUWEI MACRO 
	LOCAL NEXT
	MOV CX,12

	;CX=12,8,4
	NEXT:
		PUSH BX
		SHR BX,CL		
		MOV DL,BL
		PDL
		POP BX
		SUB CX,4
		JNZ NEXT	

	;CX=0
	AND BX,0FH
	MOV DL,BL
	PDL
	
ENDM	

	PINF INF1
	PINF INF2



	; Calculation 
	;ax,bx On behalf of the digital 
	;cl, Express al+bl Carry of 
	;ch, Express ah+bh Carry of 
	mov ax,STRING1
	mov bx,STRING2
	mov cx,0
	push ax

	add al,bl
	daa
	adc cl,0
	mov dx,ax
	
	
	pop ax
	; The carry of the lower order is required 
	add ah,cl
	add ah,bh
	mov al,ah
	daa
	adc ch,0
	mov dh,al
	; The result is  ch ,dx
	mov D,ch
	mov SUM,dx

	; Show 
	mov dl,D
	PDL
	
	mov bx,sum

	QUWEI
	
	
    	 mov ah,4ch
    	 int 21h
  



code ends
     end start

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