二进制减法-Python [英] Binary Subtraction - Python

问题描述

我想做一个二进制计算器，而减法部分有问题.这是我的代码(我尝试将其修改为我在本网站上找到的总和).

I want to make a binary calculator and I have a problem with the subtraction part. Here is my code (I have tried to adapt one for sum that I've found on this website).

    maxlen = max(len(s1), len(s2))

    s1 = s1.zfill(maxlen)
    s2 = s2.zfill(maxlen)

    result  = ''
    carry   = 0

    i = maxlen - 1
    while(i >= 0):
        s = int(s1[i]) - int(s2[i])
        if s <= 0:
            if carry == 0 and s != 0:
                carry = 1
                result = result + "1"
            else:
                result = result + "0"
        else:
            if carry == 1:
                result = result + "0"
                carry = 0   
            else:
                result = result + "1" 
        i = i - 1


    if carry>0:
        result = result + "1"

    return result[::-1]

该程序在进行某些二进制减法后可以正常工作，但在其他情况下则失败. 有人可以帮我，因为我找不到错误吗?非常感谢.

The program works fine with some binaries subtraction but it fails with others. Can someone please help me because I can't find the mistake? Thanks a lot.

推荐答案

简短答案:对于s1[i] == s2[i]和carry == 1，您的代码是错误的.

Short answer: Your code is wrong for the case when s1[i] == s2[i] and carry == 1.

更广泛的答案:您应该重组代码，使s==-1，s==0和s==1具有三个独立的大小写，然后在每种情况下分支carry的值:

Longer answer: You should restructure your code to have three separate cases for s==-1, s==0, and s==1, and then branch on the value of carry within each case:

if s == -1:  # 0-1
    if carry == 0:
        ...
    else:
        ...
elif s == 0:  # 1-1 or 0-0
    if carry == 0:
        ...
    else:
        ...
else:  # 1-0
    if carry == 0:
         ...
    else:
        ...

这样，您就每种可能性都有一个单独的障碍，因此没有机会像您初次尝试那样忽略案件.

This way you have a separate block for each possibility, so there is no chance of overlooking a case like you did on your first attempt.

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