输入输出
DATAS SEGMENT
arr dw 20 dup(?)
num db 0
string1 db 'input the number of interger:(please < 20)',13,10,'$'
string2 db 'input the interger:',13,10,'$'
string3 db 13,10,'the intergers you input is that:',13,10,'$'
DATAS ENDS
STACKS SEGMENT
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV AX,DATAS
MOV DS,AX
;输出提示语
lea dx,string1
mov ah,9
int 21h
;输入要输入数字的 个数
mov ah,1
int 21h
sub al,'0'
mov num,al
mov ah,1;这是为了清除个数之后的回车
int 21h
;输出提示语
lea dx,string2
mov ah,9
int 21h
;输入 cx存放要输入几个数
mov si,0
mov ch,0
mov cl,num
l:
call scanm;出口参数:bx
mov arr[si],bx
add si,2
loop l
;输出提示语
lea dx,string3
mov ah,9
int 21h
;输出 cx存放要输出几个数
mov ch,0
mov cl,num
mov si,0
l2:
mov ax,arr[si]
call printm;入口参数ax
;每次输出之间有个空格
mov dl,' '
mov ah,2
int 21h
;
add si,2
loop l2
MOV AH,4CH
INT 21H
;从键盘输入数->bx
scanm PROC
push ax
push cx
push dx
;暂存的数放在bx中 第一步要置0
mov bx,0
inbegin:
mov ah,1;输入一个字符到al中
int 21h
cmp al,' '
je inend
cmp al,13
je inend;如果这个数不是空格 回车 就把他加到暂存数中去
sub al,'0'
mov cl,al;新输入的数放在cl中
; bx *10 +cl
mov dl,10
mov ax,bx
mul dl; ans ax
mov bx,ax
mov ch,0
add bx,cx
jmp inbegin
inend:
pop dx
pop cx
pop ax
ret
scanm ENDP
;从 ax中取数打印到屏幕上
printm proc
push ax
push cx
push dx
;do while 除法
;dh 10
;dl cnt
mov dh,10
mov dl,0
divbegin:
div dh
push ax;ah 余数
mov ah,0
inc dl
cmp al,0
jne divbegin
mov dh,dl;dh cnt
print_begin:
cmp dh,0
je printm_end
pop cx
dec dh
mov dl,ch
add dl,'0'
mov ah,2
int 21h
jmp print_begin
printm_end:
pop dx
pop cx
pop ax
ret
printm endp
CODES ENDS
END START
输出2/16进制
DATAS SEGMENT
DATAS ENDS
STACKS SEGMENT
;此处输入堆栈段代码
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV AX,DATAS
MOV DS,AX
;ax 存要输出的数字 cl 存要输出进制 (2 or 16)
mov ax,0aaaah
mov cl,16
call print_x_radix
mov cl,2
call print_x_radix
MOV AH,4CH
INT 21H
;num->ax x_radix->cl
print_x_radix proc
push ax
push bx
push cx
push dx
;bx 存的是要与的数
;这里假设只输出2进制或者16进制
cmp cl,2
jne L0
mov cl,1
mov bx,0001h
jmp L01
L0:
mov cl,4
mov bx,000fh
L01:
;先算一下输出几次
push ax
mov ax,16
div cl
mov ch,al ;ch 要输出几次
pop ax
;移位
L1: rol ax,cl
push ax
and ax,bx
;输出ax
;分情况 如果<=9 就直接输出 否则输出字母
cmp al,9
ja L2
mov dl,al
add dl,'0'
mov ah,2
int 21h
jmp L21
L2:
mov dl,al
add dl,55
mov ah,2
int 21h
L21:
pop ax
dec ch
cmp ch,0
ja L1
cmp cl,1
jne L3
mov dl,'B'
jmp L31
L3:
mov dl,'H'
L31:
mov ah,2
int 21h
mov dl,10
mov ah,2
int 21h
pop dx
pop cx
pop bx
pop ax
ret
print_x_radix endp
CODES ENDS
END START
冒泡排序
DATAS SEGMENT
arr dw 20 dup(?)
string1 db 'input 10 nums',13,10,'$'
string2 db 'before sort',13,10,'$'
string3 db 13,10,'after sort:',13,10,'$'
DATAS ENDS
STACKS SEGMENT
;此处输入堆栈段代码
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV AX,DATAS
MOV DS,AX
;输出提示语1
lea dx,string1
mov ah,9
int 21h
;循环输入
;cx 次数 10个数
mov si,0
mov cx,10
l:
call scanm
mov arr[si],bx
add si,2
loop l
;输出提示语2
lea dx,string2
mov ah,9
int 21h
;输出没排序的
mov cx,10
mov si,0
l3:
mov ax,arr[si]
call printm
mov dl,' '
mov ah,2
int 21h
add si,2
loop l3
;排序
;bx 数组的首地址->bx 元素的个数->cl
lea bx,arr
mov cl,10
call bubble_sort
;输出提示语3
lea dx,string3
mov ah,9
int 21h
;输出结果
mov cx,10
mov si,0
l2:
mov ax,arr[si]
call printm
mov dl,' '
mov ah,2
int 21h
add si,2
loop l2
MOV AH,4CH
INT 21H
;for(int i=0;i<n-1;++i) 循环n-1次
; for(int j=0;j<n-i-1;++j)
; if(a[j]>a[j+1])
; swap(a[j],a[j+1];
;首地址->bx 数据:word 个数->cl 按照升序
bubble_sort proc
push ax
push bx
push cx
push dx
push si
push di
;dl i dh j cl n
mov dl,0
;cl ->n-1
;ch ->n-1-i
sub cl,1
i_begin:
cmp dl,cl;n-2
je i_over
mov dh,0
j_begin:
;n-2-dl
mov ch,cl
sub ch,dl
cmp dh,ch
je j_over
push dx
mov dl,dh;j->dl
mov dh,0;j->dx
mov si,dx
add si,si;si=2*j
add si,bx;si=2*j+bx
mov di,si;di=si
add di,2;di=si+2 di
pop dx
;cmp [si] [di]
mov ax,[si]
cmp ax,[di]
jna swap_over
xchg ax,[di]
mov [si],ax
swap_over:
inc dh
jmp j_begin
j_over:
inc dl
jmp i_begin
i_over:
pop di
pop si
pop dx
pop cx
pop bx
pop ax
ret
bubble_sort endp
;接受输入到 bx中
scanm PROC
; 大bug
mov bx,0
;push
push ax
push cx
push dx
inbegin:
mov ah,1
int 21h
cmp al,' '
je inend
cmp al,13
je inend
; cmp al,'9'
; ja inend
sub al,'0'
mov cl,al; cl <- al
; bx *10 +cl
mov dl,10
mov ax,bx
mul dl; ans ax
mov bx,ax
mov ch,0
add bx,cx
jmp inbegin
inend:
pop dx
pop cx
pop ax
ret
scanm ENDP
printm proc
push ax
push cx
push dx
; in:ax
;do while 除法
;dh 10
; dl 计数器 cnt
mov dh,10
mov dl,0
divbegin:
div dh
push ax;ah 余数
mov ah,0
inc dl
cmp al,0
jne divbegin
mov dh,dl;dh cnt
print_begin:
cmp dh,0
je printm_end
pop cx
dec dh
mov dl,ch
add dl,'0'
mov ah,2
int 21h
jmp print_begin
printm_end:
pop dx
pop cx
pop ax
ret
printm endp
CODES ENDS
END START
判断正数负数零
DATAS SEGMENT
;注意溢出
table db -134,98,-98,158,10,133,-45,66,-134,-122,0,-33,0,-99,0
string1 db 'positive:',13,10,'$'
string2 db 13,10,'negetive:',13,10,'$'
string3 db 13,10,'zero:',13,10,'$'
DATAS ENDS
STACKS SEGMENT
;此处输入堆栈段代码
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV AX,DATAS
MOV DS,AX
mov cx,15
mov si,0
mov al,0;正数
mov ah,0;负数
mov bl,0;0
l:
mov bh,table[si]
add bh,0
js nega_p
jz zero_p
;否则是正数
inc al
jmp myend
nega_p:
inc ah
jmp myend
zero_p:
inc bl
jmp myend
myend:
inc si
loop l
lea dx,string1
push ax
mov ah,9
int 21H
pop ax
push ax
mov ah,0
;输出al
call printm
lea dx,string2
mov ah,9
int 21H
pop ax
mov al,ah
mov ah,0
;输出ah
call printm
lea dx,string3
mov ah,9
int 21H
mov ah,0
mov al,bl
;输出bl
call printm
MOV AH,4CH
INT 21H
printm proc
push ax
push cx
push dx
; in:ax
;do while 除法
;dh 10
; dl 计数器 cnt
mov dh,10
mov dl,0
divbegin:
div dh
push ax;ah 余数
mov ah,0
inc dl
cmp al,0
jne divbegin
mov dh,dl;dh cnt
print_begin:
cmp dh,0
je printm_end
pop cx
dec dh
mov dl,ch
add dl,'0'
mov ah,2
int 21h
jmp print_begin
printm_end:
pop dx
pop cx
pop ax
ret
printm endp
CODES ENDS
END START
以有符号数输出
DATAS SEGMENT
string1 db 'input num',13,10,'$'
DATAS ENDS
STACKS SEGMENT
;此处输入堆栈段代码
STACKS ENDS
CODES SEGMENT
ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
MOV AX,DATAS
MOV DS,AX
mov bx,-2559
call printneg
MOV AH,4CH
INT 21H
printneg PROC ;bx
add bx,0
jns printneg_abolute;正数
mov dl,'-'
mov ah,2
int 21h
neg bx
printneg_abolute:
mov ax,bx
call printm
ret
printneg ENDP
printm proc
push ax
push cx
push dx
;do while 除法
;dh 10
;dl cnt
mov dh,10
mov dl,0
divbegin:
div dh
push ax;ah 余数
mov ah,0
inc dl
cmp al,0
jne divbegin
mov dh,dl;dh cnt
print_begin:
cmp dh,0
je printm_end
pop cx
dec dh
mov dl,ch
add dl,'0'
mov ah,2
int 21h
jmp print_begin
printm_end:
pop dx
pop cx
pop ax
ret
printm endp
CODES ENDS
END START
