C/C++相当于Java的doubleToRawLongBits()

Question

在Java Double.doubleToLongBits()中,实现hashCode()方法很有用.

我试图在C++中做同样的事情并编写我自己的doubleToRawLongBits()方法,因为在浏览Google之后我找不到合适的实现.

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

例如,Java Double.doubleToLongBits()为双3.94返回以下内容:

4616054510065937285

谢谢你的帮助.

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

一个简单的演员会做:

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.如果您的花车是IEE754,如果符号和指数在前面,如果CHAR_BIT == 8,你可以试试这个:

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

(在C中,(const unsigned char *)(&d)对于演员说.)

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(静态断言sizeof(double) <= sizeof(unsigned long long int)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;
}