Chapter 7 more flexible method of locating memory address
2. Program and complete the procedure in question 7.9
assume cs:codesg,ss:stacksg,ds:datasg
stacksg segment
dw 0,0,0,0,0,0,0,0
stacksg ends
datasg segment
db '1. display '
db '2. brows '
db '3. replace '
db '4. modify '
datasg ends
codesg segment
start:mov ax,stacksg
mov ss,ax
mov sp,16
mov ax,datasg
mov ds,ax
mov bx,0
mov cx,4
s0:push cx
mov si,3
mov cx,4
s:mov al,[bx+si]
and al,11011111b
mov [bx+si],al
inc si
loop s
add bx,16
pop cx
loop s0
mov ax,4c00h
int 21h
codesg ends
end start
Chapter 8 two basic problems of data processing
assume cs:codesg, ss:stack,ds:data
data segment
db '1975', '1976', '1977', '1978', '1979', '1980', '1981', '1982', '1983'
db '1984', '1985', '1986', '1987', '1988', '1989', '1990', '1991', '1992'
db '1993', '1994', '1995'
dd 16, 22, 382, 1356, 2390, 8000, 16000, 24486, 50065, 97479, 140417, 197514
dd 345980, 590827, 803530, 1183000, 1843000, 2759000, 3753000, 4649000, 5937000
dw 3, 7, 9, 13, 28, 38, 130, 220, 476, 778, 1001, 1442, 2258, 2793, 4037, 5635, 8226
dw 11542, 14430, 15257, 17800
data ends
table segment
db 21 dup ('year summ ne ?? ')
table ends
assume cs:codesg, ss:stack
stack segment
dw 8 dup (0) ;deposit cx
stack ends
codesg segment
start:mov ax,table
mov ds,ax
mov ax,stack
mov ss,ax
mov sp,16
mov ax,data
mov es,ax
mov cx,21
mov si,0
mov bx,0
year:push cx
mov di,0
mov cx,4
s:mov al,es:[si]
mov [bx+di],al
inc si
inc di
loop s
add bx,16
pop cx
loop s
mov cx,21
mov bx,0
incom:mov ax,es:[si]
mov [5+bx],ax
add si,2
mov ax,es:[si]
mov [7+bx],ax
add si,2
add bx,16
loop incom
mov cx,21
mov bx,0
staff:mov ax,es:[si]
mov [10+bx],ax
add bx,16
loop staff
mov cx,21
mov bx,0
ave:mov ax,[bx+5]
mov dx,[bx+7]
div word ptr [bx+10]
mov [13+bx],ax
add bx,16
loop ave
mov ax,4c00h
int 21h
codesg ends
end startChapter 9 principle of transfer instruction
1. Analyze the following program and think before running. Can this program return correctly?
Think again after running: why is this result?
Deepen the understanding of relevant contents through this procedure.
Can return correctly. The address of s2 is transferred to s through ax, and JMP short s actually jumps to s2. In s2, jmp short s1 calculates the offset between S1 and s2, and then adds the address of s to actually jump to the first address to complete the return
2. Programming: the string 'welcome to mask!' in green, red on green background and blue on white background will be displayed in the middle of the screen.
;In the middle of the screen, green, red, white and blue strings are displayed respectively'welcome to masm!'
assume ds:data, cs:code
data segment
db 'welcome to masm!'
data ends
code segment
start: mov ax,data
mov ds,ax
mov ax,0b800h
mov es,ax
mov bx,0
mov si,1664
mov cx,16
gree:mov al,ds:[bx]
mov ah,10000010B
mov es:[si],ax
add si,2
inc bx
loop gree
mov bx,0
mov si,1824
mov cx,16
rad:mov al,ds:[bx]
mov ah,10100100B
mov es:[si],ax
add si,2
inc bx
loop rad
mov bx,0
mov si,1984
mov cx,16
blue:mov al,ds:[bx]
mov ah,11110001B
mov es:[si],ax
add si,2
inc bx
loop blue
mov ax,4c00h
int 21h
code ends
end start
Chapter 10 CALL and RET instructions
1. Display string
problem
Display string is a function often used in real work. A general subroutine should be written to realize this function
Yes. We should provide a flexible calling interface so that the caller can decide the display position (row, column, content and color).
Subroutine description
Name: show_str
Function: display a character string ending with mouth in the specified position and color,
Parameters: (dh) - line number (value range 0 ~ 24), (d = column number (value range 0 ~ 79),
(c) - color, ds:si points to the first address of the string
Return: None
Application example: in 8 rows and 3 columns of the screen, the string in the data section is displayed in green
; In 8 rows and 3 columns of the screen, the string in the data section is displayed in green
assume cs:code,ds:data,ss:stack data segment db 'welcome to masm!',0 data ends stack segment dw 8 dup (0) stack ends code segment start:mov ax,data mov ds,ax mov ax,stack mov ss,ax mov sp,16 mov dh,8 mov dl,3 mov cl,2 call show_str mov ax,4c00h int 21h show_str:push dx push cx push ax push ss push si mov ax,0b800h mov es,ax ;Calculate the line offset. The offset is saved in bx mov al,0a0h dec dh mul dh mov bx,ax mov al,2 dec dl mul dl add bx,ax; mov di,0 mov si,0 mov al,cl s1: mov ch,0 mov cl,ds:[di] jcxz ok mov ch,al mov es:[bx+si],cx add si,2 inc di jmp short s1 ok: pop si pop ss pop ax pop cx pop dx ret code ends end
2. Solve the problem of division overflow
Name: divdw
Function: perform division operation without overflow. The dividend is dword type, the divisor is word type, and the result is dword
Parameter: (AX) = lower 16 bits of DWORD data
(DX) = upper 16 bits of DWORD data
(cx) = divisor
Return: (dx) = the upper 16 bits of the result, (ax) - the lower 16 bits of the result
(cx) = remainder
Application example: calculate 1000000 / 10 (F4240H/0A)
assume cs:code, ss:stack stack segment dw 16 dup (0) stack ends code segment start: mov ax, stack mov ss, ax mov sp, 32 mov ax, 4240h mov dx, 000fh mov cx, 0ah call divdw mov ax, 4c00h int 21h
3. Value display
Name: dtoc
Function: convert word data into a string representing decimal numbers, and the universal character string is represented by. Is the ending character.
Parameter: (ax)=word data
dssi points to the first address of the string
Return: None
Application example: programming, the data 12666 is listed in 8 rows of the screen in decimal form and displayed in green
Come on. When displaying, we call the first subroutine show str in this experiment.
assume cs:code, ss:stack stack segment dw 16 dup (0) stack ends data segment db 10 dup (0) data ends code segment start: mov ax, data mov ds, ax mov ax, stack mov ss, ax mov sp, 32 mov ax, 12666 mov si, 0 call dtoc mov dh, 8 mov dl, 3 mov cl, 2 call show_str mov ax, 4c00h int 21h dtoc: push ax push si push di push dx push bx push cx mov di, 0 mov dx, 0 mov bx, 10 devide: mov cx, ax jcxz stop div bx inc di push dx mov dx, 0 jmp devide stop: mov cx, di string: pop bx add bx, 30h mov [si], bl inc si loop string pop cx pop bx pop dx pop di pop si pop ax ret show_str: push cx push bx push ax push si push di push es ;using cx, bx, ax, si, di, es mov ax, 0b800h mov es, ax mov bx, 0 mov di, 0 mov al, 160 mul dh add bx, ax mov al, 2 mul dl add bx, ax ;bx stores address of start character mov al, cl ;al stores the color of character char: mov ch, 0 mov cl, ds:[si] jcxz zero mov ch, al mov es:[bx+di], cx add di, 2 inc si jmp char zero: pop es pop di pop si pop ax pop bx pop cx ret code ends end start
Chapter XI flag register
Write a subroutine to convert the lowercase letters in the string that contains any character and ends with 0 into large characters
Female, described below.
Name: letterc
Function: convert lowercase letters in strings ending with 0 into uppercase letters
Parameter: ds:si points to the first address of the string
assume cs:code
stack segment
dw 8 dup (0)
stack ends
data segment
db "Beginner's All-purpose Symbolic Instruction Code.", 0
data ends
code segment
begin: mov ax, stack
mov ss, ax
mov sp, 16
mov ax, data
mov ds, ax
mov si, 0
call letterc
mov ax, 4c00h
int 21h
letterc: push cx
push si
pushf
mov ch,0
s:mov cl,ds:[si]
jcxz ok
cmp cl,97
jb next
cmp cl,122
ja next
sub cl,20h
mov ds:[si],cl
next:inc si
jmp s
ok: popf
pop si
pop cx
ret
code ends
end beginChapter XII internal interruption
1. Write the program of interrupt 0, so that when the division overflow occurs, the string "divide error!" will be displayed in the middle of the screen, and then return to DOS.