更改为数字的二进制字符串整数 [英] Changing integer to binary string of digits

问题描述

我目前工作的一个模拟的MIPS处理器在C ++中的补偿结构类和有一些问题，从十进制数转换为二进制（带符号的数字两种方式）。一切都工作正常，直到最后一点，因为我现在的算法，属于出界领域​​INT 1＆LT;＆LT; = 31。只需要在正确的方向轻推让它运行起来。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;
}

我也包括我的其他算法的完整性。我的不足评论道歉，但它是相当简单的。

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;
 }

此外，如果任何人知道什么好的方法去写这些更有效地我很乐意听到这种说法。我只有两个编程入门班，但一直在玩不同的技术，看我怎么好喜欢他们的风格。

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

实际上有标准的俏皮话这些。

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();

看到这个运行作为演示。

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