C ++中的结构大小和转换为C＃

Question

I'm having an issue working from Arduino to Unity; essentially I'm building a payload struct in C++ (Arduino) and sending it to a PC using the following serial protocol:

1 byte - header

1 byte - payload size

X bytes - payload

1 byte - checksum (LRC)

The payload looks like so:

struct SamplePayload {
    double A;
    double B;
    double C;
    double D;
} payload;

when I sizeof(payload) I get 16 bytes, when I believe a double is an 8 byte data type; if I add another double the struct is 20 bytes and so on. Am I misunderstanding something? This is causing issues as this is then converted to a byte data stream and cast as a struct on receipt in C#, and I'm not sure what the equivalent datatype would be (casting as a double gives wrong values). The serial protocol in Unity also relies on the correct payload size to read out the stream.

It is probably a straightforward answer but I couldn't find it anywhere, many thanks!

Answer 1

If your application is not numerically sensitive, you could use the following approach:

Instead of using doubles within your struct (which aren't strictly standardized, as mentioned in the comments), you could use two int32_t , a and b for representing a significant and an exponent such that
a*2^b = original_double

So your struct will look something like this:

    struct SamplePayload {
    int32_t A_sig;
    int32_t A_exp;
    //B,C...
} payload;

Then on the receiving side, you will only have to multiply according to the formula above to get the original double.

C provides you with a neat function, frexp , to ease things up.
But since we store both a and b as integers, we need to modify the results a bit in order to get high precision.

Specifically, since a is guaranteed to be between 0.5 and 1, you need to multiply a by 2^30, and subtract 30 from b in order not to overflow.

Here's an example:

#include <stdio.h>      /* printf */
#include <math.h>       /* frexp */
#include <stdint.h>

int main ()
{
  double param, result;
  int32_t a,b;

  param = +235.0123123;
  result = frexp (param , &b);
  a=(result*(1<<30)) /*2^30*/; b-=30;
  printf ("%f = %d * 2^%d\n", param, a, b); //235.012312 = 985713081 * 2^-22                                                                                                                 

  return 0;
}