Question

我正在尝试编写一个代码，该代码将二进制数输入作为字符串，并且只接受1或0，否则应显示错误消息。然后它应该逐位循环以将二进制数转换为字符串为十进制。我似乎无法做到正确我有事实，它只会接受1或0的正确。但是当它进入计算时会出现一些混乱，我似乎无法将其弄清楚。目前这是我认为最接近它的工作。任何人都可以给我一些暗示或帮助我解决我的错误吗？

#include <iostream>
#include <string>
using namespace std;

string a;

int input();

int main()
{
    input();
    int decimal, x= 0, length, total = 0;
    length = a.length();
    // atempting to make it put the digits through a formula backwords.
    for (int i = length; i >= 0; i--)
    {
// Trying to make it only add the 2^x if the number is 1
        if (a[i] = '1')
        {
            //should make total equal to the old total plus 2^x if a[i] = 1
            total = total + pow(x,2);
        }
        //trying to let the power start at 0 and go up each run of the loop
        x++;
    }

    cout << endl << total;
    int stop;
    cin >> stop;
    return 0;
}

int input()
{
    int x, x2, count, repeat = 0;

    while (repeat == 0)
    {
        cout << "Enter a string representing a binary number => ";
        cin >> a;
        count = a.length();

        for (x = 0; x < count; x++)
        {

            if (a[x] != '0' && a[x] != '1')
            {
                cout << a << " is not a string representing a binary number>" << endl;
                repeat = 0;
                break;

            }
            else
                repeat = 1;



        }

    }
    return 0;
}

Answer 1

我不认为pow适合整数计算。在这种情况下，您可以使用shift operator。
a[i] = '1'将a[i]的值设置为'1'并返回'1'，这始终为真。
您不应该访问a[length]，这应该毫无意义。

固定代码：

int main()
{
    input();
    int decimal, x= 0, length, total = 0;
    length = a.length();
    // atempting to make it put the digits through a formula backwords.
    for (int i = length - 1; i >= 0; i--)
    {
// Trying to make it only add the 2^x if the number is 1
        if (a[i] == '1')
        {
            //should make total equal to the old total plus 2^x if a[i] = 1
            total = total + (1 << x);
        }
        //trying to let the power start at 0 and go up each run of the loop
        x++;
    }

    cout << endl << total;
    int stop;
    cin >> stop;
    return 0;
}

Answer 2

我会用这种方法......

{{1}}

Answer 3

我在您的代码中看到多个错误。

您的for - 循环应该从i = length - 1开始，而不是i = length。
a[i] = '1'将a[i]设置为'1'，但不进行比较。
pow(x,2)表示x^2 http://latex.codecogs.com/gif.download?x%5E2而非2^x http://latex.codecogs.com/gif.download?2%5Ex。 pow也不适用于整数运算。请改用2*2*...或1<<e。

还有更短的方法来实现它。以下是我将如何做的示例：

std::size_t fromBinaryString(const std::string &str)
{
    std::size_t result = 0;
    for (std::size_t i = 0; i < str.size(); ++i)
    {
        // '0' - '0' == 0 and '1' - '0' == 1.
        // If you don't want to assume that, you can use if or switch
        result = (result << 1) + str[i] - '0';
    }
    return result;
}

Answer 4

适用于最多32位的二进制字符串。长时间交换整数以获得64位。

#include <iostream>
#include <stdio.h>
#include <string>

using namespace std;

string getBinaryString(int value, unsigned int length, bool reverse) {
    string output = string(length, '0');
    if (!reverse) {
        for (unsigned int i = 0; i < length; i++) {
            if ((value & (1 << i)) != 0) {
                output[i] = '1';
            }
        }
    }
    else {
        for (unsigned int i = 0; i < length; i++) {
            if ((value & (1 << (length - i - 1))) != 0) {
                output[i] = '1';
            }
        }
    }
    return output;
}

unsigned long getInteger(const string& input, size_t lsbindex, size_t msbindex) {
    unsigned long val = 0;
    unsigned int offset = 0;
    if (lsbindex > msbindex) {
        size_t length = lsbindex - msbindex;
        for (size_t i = msbindex; i <= lsbindex; i++, offset++) {
            if (input[i] == '1') {
                val |= (1 << (length - offset));
            }
        }
    }
    else { //lsbindex < msbindex
        for (size_t i = lsbindex; i <= msbindex; i++, offset++) {
            if (input[i] == '1') {
                val |= (1 << offset);
            }
        }
    }
    return val;
}

int main() {
    int value = 23;
    cout << value << ": " << getBinaryString(value, 5, false) << endl;
    string str = "01011";
    cout << str << ": " << getInteger(str, 1, 3) << endl;
}

C ++二进制输入作为字符串到十进制

4 个答案: