将8个字节转换为两倍 [英] Converting 8 bytes to double

问题描述

我在将8字节转换为双精度字时遇到了一些问题。我有以下字节数组

I am facing some problem with converting 8 bytes to a double. I have following byte array

    0x98 0xf9 0x38  0x4e 0x3a 0x9f 0x1c 0x43

我正在尝试按照

     for (int i = 1; i < 8; i++)
       mult[i] = 256 * mult[i - 1];
    double out= buf[7] * mult[7] + buf[6] * mult[6] + buf[5] * mult[5] + buf[4] * mult[4] + buf[3] * mult[3] + buf[2] * mult[2] + buf[1] * mult[1] + buf[0] * mult[0];

但是它没有给出正确的答案。我出去等于4835915172658346392，实际值为2014093029293670。

But it is not giving the correct answer. I am getting out is equal to 4835915172658346392 and actual value is 2014093029293670.

注意： *（（（double *）（buf））可以正常工作，但我认为这不会对编译器和操作系统安全。

Note: *((double *) (buf)) works fine but I don't think it would be compiler and OS safe.

编辑：

long mult[8];

我从 mult [0] = 1

推荐答案

您说的是 0x98 0xf9 0x38 0x4e 0x3a 0x9f 0x1c 0x43 应该代表 2014093029293670 。

如果前者是该整数的 little-endian表示形式（采用IEEE754 binary64格式），则为true。因此，使用逐字节乘法（或等效地，移位）的方法将行不通，因为它们是算术运算。

This is true if the former is the little-endian representation of that integer in IEEE754 binary64 format. So your approach by using byte-by-byte multiplication (or equivalently, bit shifts) is not going to work, because those are arithmetic operations.

相反，您需要 alias 表示为 double 。为此，可以在一个 double 是IEEE754 binary64的小端机上进行移植：

Instead you need to alias that representation as a double. To do this portably, on a little-endian machine on which double is IEEE754 binary64:

static_assert( sizeof(double) == 8 );

double out;
memcpy(&out, buf, sizeof out);

如果您希望此代码在端序不同的计算机上运行，​​则需要重新排列 buf ，然后再执行 memcpy 。 （这是假定该表示始终以小尾数格式获得，而您并未说明）。

If you want this code to work on a machine with different endianness then you will need to rearrange buf before doing the memcpy. (This is assuming that the representation is always obtained in little-endian format, which you didn't state).

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