在两个 32 位寄存器中存储 64 位十进制(ASSEMBLY - NASM) [英] Storing a 64 bit decimal in two 32 bit registers (ASSEMBLY - NASM)

问题描述

因此，我的任务是从键盘读取最大64 位十进制数，并将该数字存储在两个 32 位寄存器(EDX:EAX).对于阅读，我必须使用一个名为 mio_readchar(它来自 mio 库)的函数，它从键盘读取一个字符，并将其存储在 AL.

So, my task is to read from the keyboard a max 64 bit decimal number, and to store that number in two 32 bit registers (EDX:EAX). For reading, I have to use a function called mio_readchar (it's from the mio library), which reads a character from the keyboard, and stores it in AL.

我不知道这是否是一项简单的任务，但我无法解决.如果你们能帮助我解决这个有趣的问题，我将不胜感激.谢谢大家！:D

I don't know whether this is an easy task or not, but I couldn't solve it. I would appreciate, if you guys could help me in solving this interesting problem. Thank you all in advance! :D

推荐答案

将其分解为更小(更简单)的部分:

Break it into smaller (simpler) pieces:

• 将当前结果设置为零

• Set the current result to zero

从键盘获取字符，检查字符是否有效(例如'0'到'9'而不是像'A'这样的疯狂的东西)，然后从中减去'0'(这样你就可以从0 到 9)

Get character from keyboard, check if character is valid (e.g. '0' to '9' and not something crazy like 'A'), then subtract '0' from it (so that you get a number from 0 to 9)

检查是否可以将当前结果乘以 10 而不会导致溢出；然后将当前结果乘以 10

Check if you can multiply the current result by 10 without causing an overflow; then multiply the current result by 10

检查是否可以将新数字添加到当前结果中而不会导致溢出；然后将新数字添加到当前结果

Check if you can add the new digit to the current result without causing an overflow; then add the new digit to the current result

跳回第二步(获取下一个字符)

Jump back to the second step (get the next character)

注意:是什么导致此循环停止(输入键?)，以及在各种错误情况下会发生什么，您需要弄清楚.

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