双精确二进制表示

Question

可能重复:
在C++中浮动到二进制

我有一个非常小的双变量,当我打印它时我得到-0.(使用C++).现在为了获得更好的精度,我尝试使用

cout.precision(18); \\i think 18 is the max precision i can get.
cout.setf(ios::fixed,ios::floatfield);
cout<<var;\\var is a double.

但它只写-0.00000000000 ...

我想看看var的确切二进制表示.

换句话说,我想看看在这个var的堆栈存储器/寄存器中写入了什么二进制数.

Answer 1

union myUnion {
    double dValue;
    uint64_t iValue;
};

myUnion myValue;
myValue.dValue=123.456;
cout << myValue.iValue;

更新:

上面的版本适用于大多数用途,但它假定64位双打.此版本不做任何假设并生成二进制表示:

    double someDouble=123.456;
    unsigned char rawBytes[sizeof(double)];

    memcpy(rawBytes,&someDouble,sizeof(double));

    //The C++ standard does not guarantee 8-bit bytes
    unsigned char startMask=1;
    while (0!=static_cast<unsigned char>(startMask<<1)) {
        startMask<<=1;
    }

    bool hasLeadBit=false;   //set this to true if you want to see leading zeros

    size_t byteIndex;
    for (byteIndex=0;byteIndex<sizeof(double);++byteIndex) {
        unsigned char bitMask=startMask;
        while (0!=bitMask) {
            if (0!=(bitMask&rawBytes[byteIndex])) {
                std::cout<<"1";
                hasLeadBit=true;
            } else if (hasLeadBit) {
                std::cout<<"0";
            }
            bitMask>>=1;
        }
    }
    if (!hasLeadBit) {
        std::cout<<"0";
    }

Answer 2

按照标准，可以保证这种方式有效：

double d = -0.0;
uint64_t u;
memcpy(&u, &d, sizeof(d));
std::cout << std::hex << u;