CODEWITHSUNDEEP
c++functionobjectprotocol-buffers
Kau
Kau

Reputation: 101

How to get information from your protobuf in C++?

Basically what I'm trying to do is get the x and y coordinates from a ProtoBuf message that is similar to the SHPObject type. I know that with SHPObject I can do things like double* x = obj->padfX and double* y = obj->padfY. However, I'm not entirely sure how to extract this same information from my ProtoBuf (code is shown below). Thank you for your help!

This what I have tried so far:

myProject::protobuf::NewShape _NewShape;

auto obj = _NewShape.shape(0);

double* x = obj.polygon(0).point(0);

Gives the error:

cannot initialize a variable of the type 'double' with an rvalue of type 'unsigned int'

And, then I tried this which compiles but doesn't do anything (does not give me the desired output):

double x_coordinate = obj.polygon(0).point(0);
double *x_ptr = &x_coordinate;

Here's my ProtoBuf file:

newShape.proto

syntax = "proto2";

package myProject.protobuf;

message NewShape {

  message Polygon 
  {
    enum PolygonType {
      POLY_TYPE_OUTER = 1;
      POLY_TYPE_INNER = 2;
    };

    optional PolygonType type = 1 [default = POLY_TYPE_OUTER];

    // x, y coordinates 
    repeated uint32 point     = 2 [packed = true];
  }

  message Shape
  {
    repeated Polygon polygon  = 1;
  }

  repeated Shape shape = 2;
}

Upvotes: 0

Views: 1053

Answers (1)

Azeem
Azeem

Reputation: 14667

Given your format, you can access points from a properly populated and deserialized object like this:

// newShape is the deserialized object here

const auto s0p0p0 = newShape.shape(0).polygon(0).point(0);
const auto s0p0p1 = newShape.shape(0).polygon(0).point(1);

Similarly, shape(1) would give you the second shape to access it points in the polygon object. You should check the *_size() methods before accessing an index to ensure valid access e.g. newShape.shape_size(), shape.polygon_size() and polygon.point_size().

If you intend to modify the message, you can use mutable_* methods to get the pointers to respective objects and then you can change those.

For example (change points of first polygon of first shape):

auto p = newShape2.mutable_shape(0);
p->mutable_polygon(0)->set_point(0, 123);
p->mutable_polygon(0)->set_point(1, 345);

Here's a complete working example of serialization and deserialization:

#include <iostream>
#include "newShape.pb.h"

int main()
{
    // Serailization

    ::myProject::protobuf::NewShape newShape1;

    auto shape1 = newShape1.add_shape();
    auto polygon1 = shape1->add_polygon();
    polygon1->add_point(1);
    polygon1->add_point(2);

    auto shape2 = newShape1.add_shape();
    auto polygon2 = shape2->add_polygon();
    polygon2->add_point(3);
    polygon2->add_point(4);

    const auto serializedData = newShape1.SerializeAsString();

    std::cout << "Serialized Data Size: " << serializedData.size() << "\n\n";

    // Send/Store Serialized Data

    // Deserialization

    ::myProject::protobuf::NewShape newShape2;

    if ( !newShape2.ParseFromString( serializedData ) )
    {
        std::cerr << "Deserialization failed!\n";
        return -1;
    }

    std::cout << "Deserialized Data Size: " << newShape2.ByteSize() << "\n\n";

    std::cout << "NewShape [Shapes: " << newShape2.shape_size() << "]\n";

    for ( int i {0}; i < newShape2.shape_size(); ++i )
    {
        std::cout << " Shape # " << i << '\n';

        const auto& shape = newShape2.shape( i );
        for ( int j {0}; j < shape.polygon_size(); ++j )
        {
            std::cout << "  Polygon # " << j << '\n';

            const auto& polygon = shape.polygon( j );
            for ( int k {0}; k < polygon.point_size(); ++k )
            {
                const auto& point = polygon.point( k );
                std::cout << "   Point # " << k << ": " << point << '\n';
            }
        }
    }

    return 0;
}

Output:

Serialized Data Size: 16

Deserialized Data Size: 16

NewShape [Shapes: 2]
 Shape # 0
  Polygon # 0
   Point # 0: 1
   Point # 1: 2
 Shape # 1
  Polygon # 0
   Point # 0: 3
   Point # 1: 4

Upvotes: 1

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