Compilation of learning - the fifteenth day

13.3-depth understanding of the int, iret and stack

Checkpoint 13.1

(1) This road I think the key is to judge

add [bp+2],bx

This is the IP address changes, i.e. jump distance forwardly, i.e. in the range bx. bx sixteen-bit register, the range from -32768 to 32767, i.e. the maximum displacement of the transfer 32768

（2）

test program:

assume cs: code 
time segment 
    db ' conversation ' , 0 
time ends 
code segment 
start: 
    mov ax, time 
    mov ds, ax 
    mov and 0 
    mov ax, 0b800h 
    mov es, ax 
    mov di, 12 * 160 
    mov bx, offset flag - flagend offset 
flag: 
    cmp byte ptr DS: [SI], 0 
    je ok 
    mov al, ds: [si] 
    mov byte ptr es: [di], al 
    mov byte ptr es: [di + 1 ] 2
    inc si
    add di,2
    int 7ch
flagend:nop
ok:
    mov ax,4c00h
    int 21h
code ends
end start

Interrupt routine:

the ASSUME CS: code 
code segment 
Start: ; program into designated idle region 
    MOV AX, CS 
    MOV DS, AX 
    MOV Si, offset do7c 
    MOV AX, 
    0 
    MOV ES, AX 
    MOV DI, 200H 
    MOV CX, offset do7cend - offset do7c 
    CLD 
    REP movsb ; set interrupt vector 
    MOV AX, 
    0 
    MOV ES, AX 
    MOV Word PTR ES: [7ch * . 4 + 2 ], 0 
    MOV Word PTR ES: [7ch * . 4 ], 200H 
    
    MOV AX, 4c00h 
    int 21H 
    
do7c: ; set IP position 
    Push BP 
    MOV BP, SP
    
    the Add SS: [BP+2],bx
    pop bp
    iret
do7cend:nop
    
code ends
end start

13.4 BIOS and DOS interrupt routines provided

Typically a program stored in the ROM on the system board, called BIOS (Basic Input Output System), BIOS mainly comprising the following parts

Hardware detection and initialization procedure
External interrupt and internal interrupts interrupt routine
A hardware device for I / O operations of the interrupt routine
And other hardware-related interrupt routine

13.5 BIOS and DOS interrupt the installation process routine

BIOS and DOS interrupt the installation process routine

After boot, the CPU a powered, initialization (CS) = 0FFFFH, (IP) = 0, automatically will be from FFFF: unit 0 begins execution, it is noted that FFFF: 0 This address is among system-wide BIOS, , so we can not go change. In FFFF: 0 Department has a jump instructions, CPU after execution of this instruction, the system will switch to perform hardware detection and initialization procedure in the BIOS
Initialization program to build BIOS supports interrupt vector, then interrupt routine entry address will be provided by the BIOS of them registered in the interrupt vector table
After the hardware detection and initialization is complete, to call int 19h to boot the operating system after the computer will be handed over to the operating system controls, where the need to pay attention to that is to go to call int 19h boot the operating system starts, so int 19h interrupt routines can not be provided by DOS
After the DOS is started, in addition to performing other work, it will also provide interrupt routines into memory, and to establish the corresponding interrupt vector

Checkpoint 13.2

(1) error, can not be changed.

(2) error, hardware should be the BIOS interrupt routine

13.6 BIOS interrupt routine application

In the display buffer B8000H ~ B8FFFFH 32KB altogether space for 80 * 25 character mode color display buffer. A screen content accounts for 4000 bytes. Display buffer a total of eight, each 4KB page.

the ASSUME CS: code 
code segment 
    ; set the cursor position 
    mov AH, 2 ; set cursor 
    mov BH, 0 ; page 0 
    mov DH, . 5 ; DH release number 
    mov DL, 12 is ; column number
     int 10H 
    
    ; displaying characters at the cursor position 
    mov AH, . 9 ; the character displayed at the cursor position 
    MOV Al, ' B ' ; characters in 
    mov bl, 11001010b; color properties 
    MOV BH, 0 ; page 0 
    MOV CX, . 3 ; the number of repeated characters
     int 10H 
    
    MOV AX, 4c00h 
    int 21H 
code ends 
End

13.7 DOS interrupt routine application

mov ah, 4ch; program returns (21h indicates interrupt subroutine calling number No. routine 4ch) 
MOV Al, 0 ; Return value
 int 21h

int 21h interrupt routine character string displayed at the cursor function

ds: dx to a string; required to display the string "$" character as the end 
MOV AH, 9     ; function 9 shows the display at the cursor strings        
 int 21H

Programming 12 shown in row 5 'Welcome to masm!'

CS the ASSUME: code 
Data segment 
    DB ' available for purchase to MASM! ' , ' $ ' 
Data ends 
code segment 
Start: 
    ; set the cursor position 
    mov AH, 2 
    mov BH, 0 
    mov DH, . 5 
    mov DL, 12 is 
    int 10H 
    
    ; display string 
    mov AX, Data 
    MOV DS, AX 
    MOV DX, 0 
    MOV AH, . 9 
    int 21H 
    
    MOV AX, 4c00h 
    int 21H 
code ends 
End Start

Experiment 13 written application interrupt routine

(1) prepare and install int 7ch interrupt routines, functions to display a character string with the end of the 0, 0 interrupt routine is mounted: at 200

Parameters: (dh) = row number, (dl) = column number, (cl) = color, ds: si points to the first address string

test program

assume cs:code
data segment
    db 'Welcome to masm!BIU',0
data ends
code segment
start:
    mov dh,12
    mov dl,30
    mov cl,2
    mov ax,data
    mov ds,ax
    mov si,0
    int 7ch
    
    mov ax,4c00h
    int 21h
code ends
end start

中断例程

assume cs:code
code segment
start:
    ;将程序写入0:200h
    mov ax,cs
    mov ds,ax
    mov si,offset func
    mov ax,0
    mov es,ax
    mov di,200h
    mov cx,offset funcend-offset func;程序长度
    cld
    rep movsb
    
    ;程序入口写入中断向量表中
    mov ax,0
    mov es,ax
    mov word ptr es:[7ch*4+2],0
    mov word ptr es:[7ch*4],200h
    
    mov ax,4c00h
    int 21h
    
func:
    push ax
    push di
    push es
    
    ;显示缓冲区地址
    mov ax,0b800h
    mov es,ax
    
    ;将已知条件转换为正确的显存地址
    mov al,160
    mul dh
    mov di,ax
    
    dec dl
    mov dh,0
    add dx,dx
    add di,dx
show:
    ;判断是否遇到末尾的0
    cmp byte ptr ds:[si],0
    je ok
    ;将显示信息写入显示缓冲区
    mov al,ds:[si]
    mov byte ptr es:[di],al
    mov es:[di+1],cl
    inc si
    add di,2
    jmp show
ok:
    ;结束
    pop es
    pop di
    pop ax
    iret
    
funcend:nop
code ends
end start

CS:IP和SS:SP在int和iret前后变化

入栈标志寄存器，CS，IP。SP=FFFAH

SS为当前段地址，CS:IP为空闲区域地址

（2）在屏幕中间显示80个'!'

测试程序

assume cs:code
code segment
start:
    mov ax,0b800h
    mov es,ax
    mov di,160*12
    mov bx,offset s-offset se
    mov cx,80
s:
    mov byte ptr es:[di],'!'
    add di,2
    int 7ch
se:
    nop
    mov ax,4c00h
    int 21h
code ends
end start

中断程序

assume cs:code
code segment
start:
    ;将程序写入0:200h
    mov ax,cs
    mov ds,ax
    mov si,offset func
    mov ax,0
    mov es,ax
    mov di,200h
    mov cx,offset funcend-offset func;程序长度
    cld
    rep movsb
    
    ;程序入口写入中断向量表中
    mov ax,0
    mov es,ax
    mov word ptr es:[7ch*4+2],0
    mov word ptr es:[7ch*4],200h
    
    mov ax,4c00h
    int 21h
    
func:
    push bp
    mov bp,sp
    dec cx
    jcxz ok
    add ss:[bp+2],bx
ok:
    pop bp
    iret
    
funcend:nop
code ends
end start

（3）下面的程序，分别在屏幕的第2,4,6,8行显示4句英文诗

assume cs:code
code segment
s1:
    db 'Good,better,best,','$'
s2:
    db 'Never let it rest,','$'
s3:
    db 'Till good is better,','$'
s4:
    db 'And better,best.','$'
s:
    dw offset s1,offset s2,offset s3,offset s4
row:
    db 2,4,6,8
    
start:
    mov ax,cs
    mov ds,ax
    mov bx,offset s
    mov si,offset row
    mov cx,4
ok:
    mov bh,0
    mov dh,ds:[si]
    mov dl,0
    mov ah,2
    int 10h
    
    mov dx,ds:[bx]
    mov ah,9
    int 21h
    inc si
    add bx,2
    loop ok
    mov ax,4c00h
    int 21h
code ends
end start