汇编程序设计入门

1. 求无符号数组BUF中的最大值存放于MAX中。（必做）

data segment
    buf db 1, 9, 23, 7, 98, 32
    n = $-buf
    max db 0
data ends

code segment
    assume cs:code, ds:data
    start:
        mov ax, data
        mov ds, ax
        mov cx, n-1
        mov si, offset buf
        mov al, [si]
        
lop:    inc si
        cmp al, [si]
        jae next
        mov al, [si]
        
next:   loop lop
        mov max, al
        mov ax, 4c00H
        int 21H
code ends
    end start  

2. 求有符号数组BUF中的最小值存放于MIN中。（必做） 

data segment
    buf db -124, -25, 123, -92, 3, 47
    n = $-buf ;$表示该行的偏移地址
    min db 0
data ends 

code segment
    assume cs:code, ds:data 
    start: 
           mov ax, data
           mov ds, ax
           ;dec buf
           ;mov bl, [buf+4]
           mov cx, n-1
           mov si, offset buf
           mov al, [si] 
              
    
    again: inc si
           cmp al, [si]
           jle next
           mov al, [si]
        
        
    next:  loop again
           mov min, al
           mov ax, 4c00H
           int 21H
           
            
        
code ends  

end start 

3. 统计数组BUF中正数、负数、0的个数顺序存放在变量NUM中。（必做）

data segment
    buf db 1, 0, 123, 9, 2, 7, -9, -4, -23, 0
    n = $-buf
    num db 3 dup(0)
data ends

code segment
    assume cs:code, ds:data

    start: mov ax, data
           mov ds, ax
           mov cx, n
           mov si, offset buf
           
    again: cmp byte ptr [si], 0
           jl a1
           jz a2
           inc num
           jmp a3
    a1:    inc num+1
           jmp a3
    a2:    inc num+2
    a3:    inc si
           loop again
           
           mov ah, 4cH
           int 21H
code ends
    end start
    

 4. 从键盘输入一个正整数，判断是否是回文数，是则输出“YES”，否则输出“NO”。例如1551、131是回文数。（选做） 

;判断回文数
data segment
    maxSize db 30H
    size db 0
    buf db 30H dup(0)
    s1 db 0aH, 'YES$'
    s2 db 0aH, 'NO$'    
data ends

code segment
    assume cs:code, ds:data
    start: mov ax, data
           mov ds, ax
           
           mov dx, offset maxSize
           mov ah, 0aH
           int 21H
           
           mov cl, size
           dec cl
           mov di, cx
           sar cx, 1
           mov bx, offset buf   
           mov si, 0
    again: mov al, [bx+si]
           cmp al, [bx+di]
           jne b2
           inc si
           dec di
           loop again
           
           mov dx, offset s1
           jmp print
    b2:    mov dx, offset s2
    print: mov ah, 09H 
           int 21H
           
           mov ah, 4cH
           int 21H
code ends
    end start 

;Debug
;mov al, [bx+si]
;mov ah, [bx+di]
;cmp al, ah

;Question
;以下均使用7654321测试
;变量名是存放数组所在存储单元的地址   
;1.mov cl, size-1 则cl为30H #
;2.cmp al, [bx+size-1]则al为37H #
;3.mov ax, [bx+6-si]当si通过inc变为1后，再执行ax变为0DH
;4.size dw 0 | mov cx, size则cx为37 07H #因为规定只保留了一个字节存储输入字符数
;5.mov al, [bx+si+[size]-1] 则al为37H
;6.mov di, word ptr size不可取 #因为size下面一块高地址是未知的

;Implement
;基址加变址寻址: bx为数组的基址，si/di表示首地址/末地址偏移量
;或者直接si/di指向首/末，个人认为不如第一种好
;问题6可用下列方法解决，可用于最下方的实现，对原来的进行替换(不可取!!!原因见问题4)
;mov [size+1], 0
;mov cx, size
;以下为不用loop的实现:
;           mov cl, size
;           dec cl
;           mov di, cx
;           sar cx, 1
;           mov bx, offset buf   
;           mov si, 0
;    again: cmp si, di
;           jae b1 ;//循环出口
;           mov al, [bx+si]
;           cmp al, [bx+di]
;           jne b2
;           inc si
;           dec di
;           jmp again
;            
;    b1:    mov dx, offset s1
;           jmp print
;    b2:    mov dx, offset s2
;    print: mov ah, 09H 
;           int 21H

5. 求出所有的水仙花数并输出。（选做）

;水仙花数
data segment
    num db 1, 0, 0
    res db 10 udp(0)
data ends

code segment
    assume ds:data, cs:code
    main proc
    start: mov ax, data
           mov ds, ax 
           mov cx, 152
           mov di, offset res
    again1:cmp cx, 160
           ja print
           mov dx, cx
           call apart
           call sum
           cmp bx, cx
           jne next
           mov [di], bx
           inc di
    next:  inc cx
           jmp again1
           
    print: mov ah, 4cH
           int 21H
           
    main endp
           
    apart proc near
        mov ax, cx
        mov bl, 100
        div bl
       
        mov [num], al
        mov al, ah
        mov ah, 0
        mov bl, 10
        div bl
       
        mov [num+1], al
        mov [num+2], ah
        ret
    apart endp
    
    sum proc near
        mov bx, 0
        mov cx, 3
        mov si, offset num
        again: mov al, [si]
               mul [si]
               mul [si]
               add bx, ax
               inc si
               loop again
        mov cx, dx
        ret     
    sum endp
                
code ends
    end start
           
 
 
;Problem
;1.cs不用进行赋值
;2.子程序写在外面会调用失败并且不停的返回start

;Implement
;因为mul会对ax进行覆写，所以需要求出三位的信息后再求立方和
;需知道调用的原理，才不会出错           

---

Reference:

https://blog.csdn.net/jia970426/article/details/85108548

https://blog.csdn.net/yang_tang/article/details/89915674

https://zhidao.baidu.com/question/412110450.html

https://blog.csdn.net/algzjh/article/details/73649546

https://blog.csdn.net/qq_40627648/article/details/84063124

https://blog.csdn.net/baidu_33836580/article/details/50579075

https://blog.csdn.net/carlyll/article/details/50437786

https://zhidao.baidu.com/question/195597064172959525.html

https://blog.csdn.net/sxhelijian/article/details/70365534

http://www.cppblog.com/luqingfei/archive/2010/08/04/122170.html

https://www.cnblogs.com/AbeDay/p/5026848.html

https://zhidao.baidu.com/question/1733497446123593347.html