在装配体x86中打印数字的三角形

剩下的是: result = 0 while (more digits to come) result *= 10; result += value of current digit这样做，82将是((0 * 10 + 8)+ 2)= 82 251将是((((0 * 10 + 2)* 10 + 5)* 10 +1 = 251 与在循环中输出数字相同，对于> 9的值，您不能简单地添加'0'并打印ascii值，您必须将其编码为ascii字符串，并显示整个字符串(就像您已经做过的一样) INT 21H/09H)也请帮自己一个忙，并消除这些问题.编写一个"decode_bin"和"encode_bin"子函数，并用对此函数的调用替换循环中的INT，否则，大约两周后您将无法读取它:-)I must do a program in tasm which have to print a triangle on numbers like this:input: n. Ex: n=4output:11 21 2 3 1 2 3 4I've managed to make my program print this thing, but i also have to make it work with numbers between 0 and 255, not only for digits. I know that i have to read a number digit by digit and create a sum like this:if i have to read 82, i read firstly 8, put it in a register, and then, when i read 2, it must be added to 8*10.Can you help me to implement this in my program?.model small.stack 100h.data msg db "Enter the desired value: $", 10, 13 nr db ?.code mov AX, @data mov DS, AX mov dl, 10 mov ah, 02h int 21h mov dl, 13 mov ah, 02h int 21h mov DX, OFFSET msg mov AH, 9 int 21h xor ax, ax mov ah, 08h ;citire prima cifra din numar int 21h mov ah, 02h mov dl, al int 21h sub al,30h mov ah,10 mul ah mov [nr],al ;mutam prima cifra inmultita cu 10 in nr mov ah, 08h int 21h mov ah, 02h mov dl, al int 21h sub al, 30h add [nr], al sub nr,30h mov dl, 10 mov ah, 02h int 21h mov dl, 13 mov ah, 02h int 21h mov cx,1 mov bx,31h mov ah, 2 mov dx, bx int 21h loop1: xor ax, ax mov al, nr cmp ax, cx je final mov dl, 10 mov ah, 02h int 21h mov dl, 13 mov ah, 02h int 21h mov bx, 0 loop2: inc bx add bx,30h mov ah, 2 mov dx, bx int 21h sub bx,30h cmp bx, cx jne loop2 inc bx add bx,30h mov ah, 2 mov dx, bx int 21h inc cx jmp loop1final: mov AH,4Ch ; Function to exit mov AL,00 ; Return 00 int 21hend 解决方案 you're half way done."if i have to read 82, i read firstly 8, put it in a register, and then, when i read 2, it must be added to 8*10"For EACH digit you read in, you have to multiply the previous value by 10. not only for the 2nd, also for the 3rd (and you can do this for the first, since the value read there was 0)what remains is: result = 0 while (more digits to come) result *= 10; result += value of current digitdoing this, 82 will be ((0*10 + 8) + 2) = 82251 will be (((0*10 + 2) *10 + 5) *10 +1 = 251same with outputting your numbers in the loop, for values > 9, you cannot simply add '0' and print the ascii value, you have to encode it as an ascii string, and display the whole string ( like you already did with INT 21H/09H )Also do yourself a favour, and divide these problems. Write a "decode_bin" and a "encode_bin" sub-functions, and replace the INTs in your loop with calls to this functions, otherwise you'll not ba able to read it, after 2 weeks or so :-) 这篇关于在装配体x86中打印数字的三角形的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！