（5）
This question will register enough, we took out a stack push way to repeatedly use register, 1,2,3-defined db here, should be "define byte", byte data
assume cs:code a segment db 1,2,3,4,5,6,7,8 a ends b segment db 1,2,3,4,5,6,7,8 b ends c segment db 0,0,0,0,0,0,0,0 c ends code segment start: mov ax, a; a segment address mov ds,ax mov ax, b; b segment address rice, Ax CX MOV, . 8 ; the number of cycles BX MOV, 0 ; offset address to exchange data AX MOV, 0 ; temporary storage + A B data s: mov al, is: [bx] add al, [bx] push ds; ds stack (which is equivalent to two push to find a place to save before ds, bx values) push bx; bx stack mov bx, c; c segment address mov ds,bx POP bx; bx removed mov [bx], al; to A + B, and assigned to the segment corresponding to the address data c pop ds; ds pop, ds paragraphs a stored address inc bx; offset + 1 loop s mov ax,4c00h int 21h code ends end start
（6）
A first data stack segment, then the stack, the position stored in the segment b
DS = 075AH, so the program entry address in 076AH, the program according to the code, we know that a first segment, segment b again, accounting for a 16-word segment, 32 bytes (1F), so 076A: 0H ~ 076A: 1FH is a data segment, segment b, 8 accounting for 16 bytes, in paragraph b 076C: 0H ~ 076C: FH
assume cs:code a segment dw 1,2,3,4,5,6,7,8,9,0ah,0bh,0ch,0dh,0eh,0fh,0ffh a ends b segment dw 0,0,0,0,0,0,0,0 b ends code segment start: mov ax,a mov ds,ax mov ax,b rice, Ax mov bx,0 mov cx,8 s0: mov ax,[bx] push ax add bx,2 loop s0 mov bx,0 s1: pop ax es: [bx], Ax add bx,2 loop s1 mov ax,4c00h int 21h code ends end start
Chapter VII of the more flexible method of locating memory address
It is strongly recommended to go to 8.1 to 8.4 to read, because you write the code in this chapter step on the pit about [...] of 8.1 to 8.4 are explained in
7.1 and and or instructions
(1) and command: command logic and performs a bitwise
mov al,01100011B
and al,00111011B
(2) or instruction ...
mov al,01100011B
or al,00111011B
About 7.2 ASCII code
7.3 Data given in characters
7.4 case conversion issues
assume cs:codesg,ds:datasg datasg segment db 'BaSiC' db 'iNfOrMaTiOn' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov cx,5 mov bx,0 mov al,11011111b s: and ds:[bx],al inc bx loop s mov cx,11 mov bx,0 mov al,00100000b s0: or ds:[bx],al inc bx loop s0 mov ax,4c00h int 21h codesg ends end start
7.5 [bx+idata]
instruction
mov ax,[bx,200] ;(ax) = ( (ds) * 16 + (bx) + 200)
equal
mov ax,[200+bx]
mov ax,200[bx]
mov ax,[bx].200
Question 7.1
(Ax) = 00B
(bx)=1000h
(cx)=0606h
Array processing with 7.6 [bx + idata] manner
assume cs:codesg,ds:datasg datasg segment db 'BaSiC' db 'MinIX' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov cx,5 mov bx,0 s: mov al,11011111b and ds:[bx],al mov al,00100000b or ds:5[bx],al inc bx loop s mov ax,4c00h int 21h codesg ends end start
7.7 GB 和 DI
8086CPU si and di are functionally similar and bx si and di registers can be divided into two 8-bit registers
mov bx,0 mov ax,[bx] purple and 0 mov ax, [you] mov on, 0 mov ax, [in]
Issue 7.2
Code book:
assume cs:codesg,ds:datasg datasg segment db 'welcome to masm!' db '................' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov a, 0 mov di, 16 mov cx,8 p: mov ax, [a] mov [in], ax add si, 2 add on, 2 loop s mov ax,4c00h int 21h codesg ends end start
written:
assume cs:codesg,ds:datasg datasg segment db 'welcome to masm!' db '................' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov si,0 mov cx,16 s: mov al,[si] mov [si+16],al inc si loop s mov ax,4c00h int 21h codesg ends end start
不能直接给段寄存器赋值，要使用普通寄存器，不能给段存储器:[偏移地址]使用直接赋值或者段寄存器赋值，这和前面的段寄存器一样的。
问题 7.3
和自己写的7.2一样
7.8 [bx+si] 和 [bx+di]
[bx+si]也可以写成[bx][si]
mov ax,[bx+si]
mov ax,[bx][si]
问题 7.4
(ax)=00BEH
(bx)=1000H
(cx)=0606H
7.9 [bx+si+idata] 和 [bx+di+idata]
mov ax,[bx+200+si] mov ax,[200+bx+si] mov ax,200[bx][si] mov ax,[bx].200[si] mov ax,[bx][si].200
问题 7.5
ax=0006h
cx=6a00h
bx=226ah
7.10 不同的寻址方式的灵活应用
问题 7.6
开头字母都在偏移地址3
assume cs:codesg,ds:datasg datasg segment db '1. file ' db '2. edit ' db '3. search ' db '4. view ' db '5. options ' db '6. help ' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov si,3 mov cx,6 mov al,11011111b s: and [si],al add si,10h loop s mov ax,4c00h int 21h codesg ends end start
问题 7.7
assume cs:codesg,ds:datasg datasg segment db 'ibm ' db 'dec ' db 'var ' db 'vec ' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov si,0 mov cx,4 s0: mov dx,cx mov bx,0 mov cx,3 s: mov al,11011111b and [bx+si],al inc bx loop s add si,10h mov cx,dx loop s0 mov ax,4c00h int 21h codesg ends end start
在上面的程序中，使用dx来临时储存外层循环cx的值，如果寄存器不够怎么办？两种方法
1.入栈出栈
assume cs:codesg,ds:datasg datasg segment db 'ibm ' db 'dec ' db 'var ' db 'vec ' datasg ends codesg segment start: mov ax,datasg mov ds,ax mov si,0 mov cx,4 s0: push cx mov bx,0 mov cx,3 s: mov al,11011111b and [bx+si],al inc bx loop s add si,10h pop cx loop s0 mov ax,4c00h int 21h codesg ends end start
上面这个是自己写的，虽然能够得出正确答案，但是最好定一个栈空间
assume cs:codesg,ds:datasg,ss:stacksg datasg segment db 'ibm ' db 'dec ' db 'var ' db 'vec ' datasg ends stacksg segment dw 0,0,0,0,0,0,0,0 stacksg ends codesg segment start: mov ax,datasg mov ds,ax mov ax,stacksg mov sp,15 mov ss,ax mov si,0 mov cx,4 s0: push cx mov bx,0 mov cx,3 s: mov al,11011111b and [bx+si],al inc bx loop s add si,10h pop cx loop s0 mov ax,4c00h int 21h codesg ends end start
第二种，cx放入内存空间
assume cs:codesg,ds:datasg datasg segment db 'ibm ' db 'dec ' db 'var ' db 'vec ' dw 0 ;有段空间才能使用 datasg ends codesg segment start: mov ax,datasg mov ds,ax mov si,0 mov cx,4 s0: mov ds:[40h],cx ;必须加上ds mov bx,0 mov cx,3 s: mov al,11011111b and [bx+si],al inc bx loop s add si,10h mov cx,ds:[40h] ;必须加上ds loop s0 mov ax,4c00h int 21h codesg ends end start
为什么在[40h]必须加上ds,不然运行失败？
因为在这里，我们使用另一个段寄存器来储存cx数据，而使用[40h]默认是ds段寄存器的地址，但我们储存cx的段寄存器不一定是ds,可以命名其他的，所以需要加上。（见 8.3-(3) ））
问题 7.9
SI源变址寄存器，DI目地变址寄存器，两者不能同时在[...]中使用，具体看点击1--点击2
assume cs:codesg,ds:codesg datasg segment db '1. display ' db '2. brows ' db '3. replace ' db '4. modify ' dw 0 datasg ends codesg segment start: mov ax,datasg mov ds,ax mov bx,0 mov cx,4 s0: mov ds:[40],cx mov di,3 ;注意开始位置 mov cx,4 s: mov al,11011111b and [bx+di],al inc di loop s add bx,10h mov cx,ds:[40h] loop s0 mov ax,4c00h int 21h codesg ends end start
实验 6 实践课程中的程序
前面已做
第八章 数据处理的两个基本问题
reg(寄存器)：ax, bx, cx, dx, ah, al, bh, bl, ch, cl, dh, dl, sp, bp, si, di;
sreg(段寄存器): ds, ss, cs, es;
8.1 bx，si，di和bp
https://www.cnblogs.com/Mayfly-nymph/p/11079189.html
8.2 机器指令处理的数据在什么地方
机器指令进行数据处理：读取，写入，运算
指令执行前，数据可存在于：CPU内部，内存，端口
| 机器码 | 汇编指令 | 指令执行前数据的位置 |
| 8E1E0000 | mov bx,[0] | 内存，ds:0单元 |
| 89C3 | mov bx,ax | CPU内部，ax寄存器 |
| BB0100 | mov bx,1 | CPU内部，指令缓冲器 |
8.3 汇编语言中数据位置的表达
（1）立即数
直接包含在机器指令的数据（执行前在CPU的指令缓冲器）
mov ax,1
add ax,5
or ax,00100000b
mov bl,'a'
（2）寄存器
指令要处理的数据在寄存器中，也就是给出寄存器
mov ax,bx
mov ds,ax
push bx
mov ds:[0],bx
push ds
mov ss,ax
mov sp,ax
（3）段地址(SA)和偏移地址(EA)
指令要处理在内存中的数据，可以用[X]的指令给出EA，SA在某个段寄存器中
段地址默认在ds中的
mov ax,[0]
mov ax,[di]
mov ax,[bx+di+8]
段地址默认在ss中的
mov ax,[bp]
mov ax,[bp+8]
mov ax,[bp+si+8]
显性给出段寄存器地址
mov ax,ds:[bp]
mov ax,es:[bx]
mov ax,ss:[bx+si]
mov ax,cs:[bx+si+8]