十进制转换为N进制问题

问题描述：给定十进制数(非负的实数)的字符串表示与转换的进制，转换的精度，要求输出转换后的进制数的字符表示

代码：

char unsignedToChar(unsigned n)
{
    char ch = '#';

    if (0 <= n && n <= 9)
        ch = '0' + n;
    else
    if (10 <= n && n <= 36)
        ch = n - 10 + 'A';

    return ch;
}

unsigned charToUnsigned(char ch)
{
    unsigned n = 0;

    if ('0' <= ch && ch <= '9')
        n = ch - '0';
    else
    if ('A' <= ch && ch <= 'Z')
        n = ch - 'A' + 10;
    return n;
}

std::string decimalTo(const std::string &decimal, unsigned digit, unsigned accuracy) 
{
    unsigned accuracyCopy = accuracy;
    std::stringstream strToDecimal(decimal);
    std::stack<char> bInt;
    std::string decNumStr;
    unsigned integer;
    double decNum, decimalPlace;
    const unsigned digitNum = digit;

    /* 字符串流将十进制字符串表示转换为double类型 */
    strToDecimal >> decNum;
    /* 获取十进制数的整数与小数部分 */
    integer = static_cast<unsigned>(decNum / 1);
    decimalPlace = decNum - integer;

    /* 十进制整数部分转换为二进制 */
    while (integer)
    {
        bInt.push(unsignedToChar(integer % digitNum));
        integer /= digitNum;
    }
    while (!bInt.empty())
    {
        decNumStr.push_back(bInt.top());
        bInt.pop();
    }

    /* 十进制小数部分转换为二进制 */
    if (accuracyCopy != 0)
        decNumStr.push_back('.');
    while (decimalPlace && accuracyCopy != 0)
    {
        --accuracyCopy;
        decimalPlace *= digitNum;
        decNumStr.push_back(unsignedToChar(static_cast<char>(decimalPlace / 1)));
        decimalPlace -= static_cast<int>(decimalPlace) / 1;
    }
    while (accuracyCopy != 0)
    {
        --accuracyCopy;
        decNumStr.push_back(unsignedToChar(0));
    }
    return decNumStr;
}</span>