Converting struct to byte and back to struct

Question

I'm currently working with Arduino Unos, 9DOFs, and XBees, and I was trying to create a struct that could be sent over serial, byte by byte, and then re-constructed into a struct.

So far I have the following code:

struct AMG_ANGLES {
    float yaw;
    float pitch;
    float roll;
};

int main() {
    AMG_ANGLES struct_data;

    struct_data.yaw = 87.96;
    struct_data.pitch = -114.58;
    struct_data.roll = 100.50;

    char* data = new char[sizeof(struct_data)];

    for(unsigned int i = 0; i<sizeof(struct_data); i++){
        // cout << (char*)(&struct_data+i) << endl;
        data[i] = (char*)(&struct_data+i); //Store the bytes of the struct to an array.
    }

    AMG_ANGLES* tmp = (AMG_ANGLES*)data; //Re-make the struct
    cout << tmp.yaw; //Display the yaw to see if it's correct.
}

Source: http://codepad.org/xMgxGY9Q

This code doesn't seem to work, and I'm not sure what I'm doing wrong.

How do I solve this?

a plead to all c++ programmers - use c++ style casts instead of c-style casts. — Tobias Langner, Dec 08 '12 at 09:06
Didn't realize they were c-style casts as this is what my teachers taught my in my C++ class — Steven10172, Dec 08 '12 at 09:09

score 40 · Accepted Answer · edited Dec 08 '12 at 21:28

40

It seems I've solved my issue with the following code.

struct AMG_ANGLES {
    float yaw;
    float pitch;
    float roll;
};

int main() {
    AMG_ANGLES struct_data;

    struct_data.yaw = 87.96;
    struct_data.pitch = -114.58;
    struct_data.roll = 100.50;

    //Sending Side
    char b[sizeof(struct_data)];
    memcpy(b, &struct_data, sizeof(struct_data));

    //Receiving Side
    AMG_ANGLES tmp; //Re-make the struct
    memcpy(&tmp, b, sizeof(tmp));
    cout << tmp.yaw; //Display the yaw to see if it's correct
}

WARNING: This code will only work if sending and receiving are using the same endian architecture.

edited Dec 08 '12 at 21:28

Peter Mortensen

30,738
21
105
131

answered Dec 08 '12 at 08:56

Steven10172

2,003
4
21
37

2

An addition to your warning, both sides need to represent a float identically as well. I'd recommend sending it as fixed point integer (in your example above: 8796, -11458, 10050) and document the endianness of the values. Use functions like `htobe32()` from `endian.h` to convert between host and big- or little-endian. – tomlogic Dec 10 '12 at 17:29
3

@Steven10172 Couldn't the struct have padding? This would not tightly pack the structure. – Trevor Hickey Dec 15 '15 at 13:44

score 7 · Answer 2 · answered Dec 08 '12 at 08:48

7

You do things in the wrong order, the expression

&struct_data+i

takes the address of struct_data and increases it by i times the size of the structure.

Try this instead:

*((char *) &struct_data + i)

This converts the address of struct_data to a char * and then adds the index, and then uses the dereference operator (unary *) to get the "char" at that address.

answered Dec 08 '12 at 08:48

Some programmer dude

400,186
35
402
621

It doesn't seem to let me call it. I get "Line 23: error: request for member 'yaw' in 'tmp', which is of non-class type 'AMG_ANGLES*'" – Steven10172 Dec 08 '12 at 08:53
See my answer. You have to use '->' – jtepe Dec 08 '12 at 08:54
This has helped a lot right now, 6 years later. Thank you kind sir! – Valdrinium Mar 10 '18 at 19:26

score 4 · Answer 3 · edited May 10 '16 at 04:22

4

Always utilize data structures to its fullest..

union AMG_ANGLES {
  struct {
    float yaw;
    float pitch;
    float roll;
  }data;
  char  size8[3*8];
  int   size32[3*4];
  float size64[3*1];
};

edited May 10 '16 at 04:22

Box Box Box Box

5,094
10
49
67

answered May 09 '16 at 20:01

CharLess

49
1
1

Why is float 8 bytes long? – Alin I Nov 29 '16 at 00:21
it's not. float is 32bit. But there is double, which is basically just a 64bit float. Also there is long double. – Simon Nitzsche Nov 03 '17 at 05:15

score 1 · Answer 4 · answered Dec 08 '12 at 08:53

1

for(unsigned int i = 0; i<sizeof(struct_data); i++){
    // +i has to be outside of the parentheses in order to increment the address
    // by the size of a char. Otherwise you would increment by the size of
    // struct_data. You also have to dereference the whole thing, or you will
    // assign an address to data[i]
    data[i] = *((char*)(&struct_data) + i); 
}

AMG_ANGLES* tmp = (AMG_ANGLES*)data; //Re-Make the struct
//tmp is a pointer so you have to use -> which is shorthand for (*tmp).yaw
cout << tmp->yaw; 
}

answered Dec 08 '12 at 08:53

jtepe

3,242
2
24
31

What solution would be more efficient? Yours or mine? This operation will be running ~30-50 times a second – Steven10172 Dec 08 '12 at 08:58
memcpy is probably more efficient. So, I would go with your solution. It also makes the code easier to read. – jtepe Dec 08 '12 at 09:04
Thanks for explaining why I would have to use -> I've still new to C++, only started to learn 2-3months ago, and all the *'s and &'s still confuse me sometimes – Steven10172 Dec 08 '12 at 09:07
yup, pointers are a classic hurdle C and C++ beginners have to clear. As to why memcpy is more efficient, have a look at [this question](http://stackoverflow.com/questions/1209529/optimized-memcpy) – jtepe Dec 08 '12 at 09:11

Converting struct to byte and back to struct

4 Answers4

Linked