输入一个数N，对1到N的所有自然数求和

程序运行：

 

代码：

datas segmentN_string_max_length      db 0ffhN_string                 db  0, 100h dup(?)N                        dw 0 sum                      dd 0 input              db 'input N=$'output             db 0dh,0ah,'output sum=$'datas endsstacks segment stackdb  100h dup(?)stacks endscodes segmentassume cs:codes,ds:datas,ss:stacks
main   proc  far
start:push dsmov ax,0hpush axmov ax,datas          ;初始化dsmov ds,ax;输入提示lea dx,inputmov ah,9int 21h;输入Nlea dx,N_string_max_lengthmov ah,10 int 21h;转换成为数字lea si,N_stringcall todigitmov N,axmov ax,0 mov dx,0mov cx,N	           ;循环的次数mov bx,1 	       ;求和从一开始s:add ax,bx        ;求和adc dx,0 inc bx loop smov word ptr sum,ax         ;保存求和的结果mov word ptr sum+2,dxlea dx,output               ;输出提示mov ah,9int 21h mov ax,word ptr sum        ;输出求和mov dx,word ptr sum+2call print_decimalretmain endptodigit proc near    uses bx cx dx si di ;si ,axmov cl,[si]      ;设置循环次数xor ch,ch inc si           ;si指向第一个元素mov ax,0         ;初始化ax,dxmov dx,0 mov di,10        ;除数dig:mov bl,[si]     cmp bl,'0'       ;判断是否小于'0'jb err           ;若低于'0'，则错误字符cmp bl,'9'       ;判断是否大于'9'ja err           ;若高于'9'，则错误字符sub bl,30h       ;否则，将数字字符转成数字xor bh,bh       mul di           ;乘十add ax,bx        ;加上个位adc dx,0          inc siloop digcmp dx,0          ;判断高位是否为0jnz err           ;若不为0,错误jmp exit           err:               ;输出错误push ds push cs pop ds lea dx,error_tip			mov ah,9int 21hpop dsmov ah,4ch		int 21hexit:ret error_tip  db 0dh,0ah,'input error$'
todigit endpprint_decimal proc near    uses ax  bx cx dx mov bx,10 mov cx,0      dea:      div bx push dx inc cx xor dx,dx cmp ax,0 jnz deadea1:pop dxadd dl,30hmov ah,2int 21hloop dea1		ret 
print_decimal endp 
codes endsend main