Question

在Java中Double.doubleToLongBits()对于实现hashCode()方法非常有用。

我正在尝试在C ++中执行相同的操作并编写自己的doubleToRawLongBits()方法，因为在浏览Google之后我无法找到合适的实现。

我可以从std::frexp(numbr,&exp)获取signif和exponent并且可以确定符号，但无法弄清楚使用按位运算符来获得Java等价物。

例如，Java的Double.doubleToLongBits()返回以下双3.94：

4616054510065937285

感谢您的帮助。

格雷厄姆

以下是从Double.doubleToRawLongBits（）

复制和粘贴的文档

===Java Double.doubleToRawLongBits() description===

/**
     * Returns a representation of the specified floating-point value
     * according to the IEEE 754 floating-point "double
     * format" bit layout, preserving Not-a-Number (NaN) values.
     * <p>
     * Bit 63 (the bit that is selected by the mask 
     * <code>0x8000000000000000L</code>) represents the sign of the 
     * floating-point number. Bits 
     * 62-52 (the bits that are selected by the mask 
     * <code>0x7ff0000000000000L</code>) represent the exponent. Bits 51-0 
     * (the bits that are selected by the mask 
     * <code>0x000fffffffffffffL</code>) represent the significand 
     * (sometimes called the mantissa) of the floating-point number. 
     * <p>
     * If the argument is positive infinity, the result is
     * <code>0x7ff0000000000000L</code>.
     * <p>
     * If the argument is negative infinity, the result is
     * <code>0xfff0000000000000L</code>.
     * <p>
     * If the argument is NaN, the result is the <code>long</code>
     * integer representing the actual NaN value.  Unlike the
     * <code>doubleToLongBits</code> method,
     * <code>doubleToRawLongBits</code> does not collapse all the bit
     * patterns encoding a NaN to a single &quot;canonical&quot; NaN
     * value.
     * <p>
     * In all cases, the result is a <code>long</code> integer that,
     * when given to the {@link #longBitsToDouble(long)} method, will
     * produce a floating-point value the same as the argument to
     * <code>doubleToRawLongBits</code>.
     *
     * @param   value   a <code>double</code> precision floating-point number.
     * @return the bits that represent the floating-point number.
     * @since 1.3
     */
    public static native long doubleToRawLongBits(double value);

Answer 1

简单演员会做：

double d = 0.5;

const unsigned char * buf = reinterpret_cast<const unsigned char *>(&d);

for (unsigned int i = 0; i != sizeof(double); ++i)
  std::printf("The byte at position %u is 0x%02X.\n", i, buf[i]);

符号位和指数位取决于您的平台和字节顺序。如果您的浮点数是IEE754，如果符号和指数位于前面且CHAR_BIT == 8，则可以尝试：

const bool sign = buf[0] & 0x80;
const int exponent = ((buf[0] & 0x7F) << 4) + (buf[1] >> 4) - 1023;

（在C中，为演员说(const unsigned char *)(&d)。）

更新：要创建具有相同位的整数，必须先使用整数然后复制：

unsigned long long int u;
unsigned char * pu = reinterpret_cast<unsigned char *>(&u);
std::copy(buf, buf + sizeof(double), pu);

为此，你必须记住几件事：整数的大小必须足够（sizeof(double) <= sizeof(unsigned long long int)的静态断言应该做的伎俩），如果整数实际上更大，那么你'只是复制到它的一部分。我相信你会明白这一点，但是:-)（你可以使用一些模板魔术来创建一个正确大小的整数，如果你真的想要的话。）

Answer 2

#include <stdint.h>

static inline uint64_t doubleToRawBits(double x) {
    uint64_t bits;
    memcpy(&bits, &x, sizeof bits);
    return bits;
}

Answer 3

我喜欢这些事情的工会。

union double_and_buffer {
    double d;
    unsigned char byte_buff[ sizeof(double) ];
} dab;

dab.d = 1.0;
for ( int i = 0; i < sizeof(dab.byte_buff); ++i )
{
    cout << hex byte_buff[ i ];
}

我认为它会让你更清楚自己在做什么，并让编译器完成所有数学运算。

C / C ++相当于Java的doubleToRawLongBits（）

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