将整数更改为二进制数字字符串 [英] Changing integer to binary string of digits

问题描述

我目前正在用 C++ 模拟 MIPS 处理器，用于一个 comp 架构类，并且在从十进制数转换为二进制数(双向有符号数)时遇到了一些问题.直到最后一点，一切都运行良好，因为我当前的算法在 1<<=31 上属于 int 的边界区域.只需要朝正确的方向轻推即可启动并运行.谢谢！

I'm currently working on a simulation of the MIPS processor in C++ for a comp architecture class and having some problems converting from decimal numbers to binary (signed numbers both ways). Everything's working fine until the very last bit because my current algorithm falls into out of bounds areas for int on 1<<=31. Just need a nudge in the right direction to get it up and running. Thanks!

//Assume 32 bit decimal number
string DecimalToBinaryString(int a)
{
    string binary = "";
    int mask = 1;
    for(int i = 0; i < 31; i++)
    {
        if((mask&a) >= 1)
            binary = "1"+binary;
        else
            binary = "0"+binary;
        mask<<=1;
    }
    cout<<binary<<endl;
    return binary;
}

为了完整性，我还包括了我的其他算法.对于没有发表评论，我深表歉意，但这是相当直截了当的.

I'm also including my other algorithm for completeness. I apologize for the lack of comments, but it's fairly straight forward.

int BinaryStringToDecimal(string a)
{
    int num = 0;
    bool neg = false;
    if(a.at(0) == '1')
    {
        neg = true;
        for(int x = a.length()-1; x >= 0; x--)
        {
            if(a.at(x) == '1')
                a.at(x) = '0';
            else a.at(x) = '1';
        }
        a.at(a.length()-1) += 1;
        for(int x = a.length()-1; x >= 0; x--)
        {
            if(a.at(x) == '2')
            {
                if(x-1 >= 0)
                {
                    if(a.at(x-1) == '1')
                        a.at(x-1) = '2';
                    if(a.at(x-1) == '0')
                        a.at(x-1) = '1';
                    a.at(x) = '0';
                }
            }
            else if(a.at(x) == '3')
            {
                if(x-1 >= 0)
                    a.at(x-1) += '2';
                a.at(x) = '1';
            }
        }
        if(a.at(0) == '2')
            a.at(0) = '0';
        else if(a.at(0) == '3')
            a.at(0) = '1';
    }
    for(int x = a.length()-1; x >= 0; x--)
    {
        if(a.at(x) == '1')
            num += pow(2.0, a.length()-x-1);
    }
    if(neg)
        num = num*-1;   
    return num;
 }

另外，如果有人知道更有效地编写这些内容的任何好方法，我很想听听.我只上过两门编程入门课程，但一直在尝试不同的技术，看看我有多喜欢他们的风格.

Also if anyone knows any good ways to go about writing these more efficiently I'd love to hear it. I've only had the two introductory programming classes but have been playing with different techniques to see how well I like their style.

推荐答案

实际上有这些标准的单行代码.

There are actually standard one-liners for these.

#include <bitset>

std::string s = std::bitset< 64 >( 12345 ).to_string(); // string conversion

std::cout << std::bitset< 64 >( 54321 ) << ' '; // direct output

std::bitset< 64 > input;
std::cin >> input;
unsigned long ul = input.to_ulong();

将此运行视为演示.

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