在重新解释为浮点数时，我可以指定字节顺序吗？

I am writing a C program that communicates over modbus TCP. I send the float 3.14 on my modbus client, with a hexadecimal representation of 0x4048F5C3. My C code is correctly receiving the modbus packet, and now I just need to reinterpret the unsigned char *buffer as a float so I can get the value 3.14. Here is my code

我正在编写一个通过modbus TCP进行通信的C程序。我在我的modbus客户端发送浮点数3.14，十六进制表示为0x4048F5C3。我的C代码正确接收modbus数据包，现在我只需要将unsigned char * buffer重新解释为float，这样我就可以得到值3.14。这是我的代码

int main()
{
    setup_modbus_tcp();
    unsigned char buffer[4] = { 0x00, 0x00, 0x00, 0x00 };
    get_modbus_data(buffer, 4);
        //Buffer now holds { 0x40, 0x48, 0xF5, 0xC3 };
    float value = *(float *)buffer;
        //value is now -490.564453
}

Now, I can tell that this problem is because of endianness, because if I swap endianness so that the buffer looks like this:

现在，我可以说这个问题是因为字节序，因为如果我交换字节序以便缓冲区看起来像这样：

{ 0xC3, 0xF5, 0x48, 0x40 };

then everything works fine. But this feels a little bit messy. Do I have to swap endianness? Or is there a way that I can specify endianness when I do:

一切都很好。但这感觉有点乱。我必须交换字节序吗？或者有一种方法可以在我执行时指定字节顺序：

float value = *(float *)buffer;

1 个解决方案

#1

The buffer must be handled manually, this includes changes to endianness. A function can always be used that converts the buffer to a floating point format.

必须手动处理缓冲区，这包括对字节序的更改。始终可以使用将缓冲区转换为浮点格式的函数。

float value = Convert( buffer );

To add, you way of converting the buffer is incorrect:

要添加，您转换缓冲区的方式不正确：

float value = *(float *)buffer;

An array of unsigned chars cannot be simply to a float type, as it causes undefined behavior. The memory must be copied to an object of type float:

无符号字符数组不能简单地为float类型，因为它会导致未定义的行为。必须将内存复制到float类型的对象：

float value;
memcpy( &value , buffer , sizeof( value ) );  //assumes endianness has already been handled

#1