将8个字节转换为double值[重复]

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How to convert a byte array into double in C? 3 answers

如何在C中将字节数组转换为double? 3个答案

I read from a socket some hex data. The specification for this protocol tells me that the following 8 bytes represent a double Value.

我从套接字读取一些十六进制数据。该协议的规范告诉我,以下8个字节代表双值。

So I have for example 8 byte like:

所以我有例如8字节:

0x3F 0xD1 0x9B 0x94 0xC0 0x00 0x00 0x00

(This value is saved in a char array so Array[0] = 0x3F, `Array[1] = 0xD1è...)

(此值保存在char数组中,因此Array [0] = 0x3F,`Array [1] =0xD1è...)

The represented double value is: 0.275120913982

代表的双重值是:0.275120913982

How can I convert these 8 bytes to this double value?

如何将这8个字节转换为此double值?

I tried a lot of different things, but nothing works really. So I have no idea how I can manipulate it to a double.

我尝试了很多不同的东西,但没有任何作用。所以我不知道如何将它操纵为双倍。

1 个解决方案

#1

You can use union, this works for me:

你可以使用union,这对我有用:

#include <iostream>
using namespace std;

int main() {
    union { char b[8]; double d; };
    b[7] = 0x3F;
    b[6] = 0xD1;
    b[5] = 0x9B;
    b[4] = 0x94;
    b[3] = 0xC0;
    b[2] = 0x00;
    b[1] = 0x00;
    b[0] = 0x00;

    cout << "Double: " << d << endl;

    return 0;
}

Please note reverse order of bytes. It depends on endianness and can differ on your machine. Anyway it outputs:

请注意字节的反向顺序。它取决于字节顺序,并且可能在您的计算机上有所不同。无论如何它输出:

Double: 0.275121

#1