How do I read a binary file in C++ if I generate it in Java?

Question

I'm working on an encryption algorithm and I need to generate some information in Java (a binary file) to read in C++.

I'm not sure if the problem is how I create the binary file or how I read it, though I can perfectly read the information in Java.

So I made a simple test. In Java I save the number 9 to a binary file, and then I try to read it in C++, but it does not read a number.

Can someone please tell me how I can do this?

The Java code:

int x = 9;

        try{
            ObjectOutputStream salida=new ObjectOutputStream(new 
            FileOutputStream("test.bin"));
            salida.writeInt(x);
            salida.close();
            System.out.println("saved");
        } catch(Exception e){
            System.out.println(e);
        }

The C++ code:

streampos size;
  char * memblock;

  ifstream file ("test.bin", ios::in|ios::binary|ios::ate);
  if (file.is_open())
  {
    size = file.tellg();
    cout<< size << endl;
    memblock = new char [size];
    file.seekg (0, ios::beg);
    file.read (memblock, size);
    file.close();

    int i;
    for (i = 0; i < sizeof(memblock); i++)
    {
        cout << memblock[i] <<endl;
    }
 delete[] memblock;
  }
  else cout << "Unable to open file";

This is the output:

�
�


w

Answer 1

Your problem is that you are using ObjectOutputStream to write the data. This encodes the object graph in a Java-specific form intended to be read with ObjectInputStream . To make the data stream compatible with C++ you would need to do one of two things:

1. Implement in C++ code that understands the output format produced by ObjectOutputStream -- ie re-implement in C++ what Java does in ObjectInputStream . This is NOT recommended.
2. Write your data out from Java using a standard FileOutputStream , in a serialized format that you define, that then can be read by your C++ code. How you specify and implement this is up to you but can be very simple, depending on the complexity of your data.

Answer 2

Yor write to file int (4 bytes?), in your file in hex must bu such data as 09 00 00 00 . In your cpp code you read it to char array (you read bytes!), file.tellg(); return 4, and you read to char * memblock array {9, 0, 0, 0}. And after it you print it as chars cout << memblock[i] <<endl; . So you can print your out array as

for (i = 0; i < size / sizeof(int) / ; i++) {
   cout << ((int*)memblock)[i] <<endl;
}

or read it in int array

int* memblock;
...
memblock = new int[size / sizeof(int)];
...
file.read (memblock, (size / sizeof(int)) * sizeof(int));

Answer 3

Java code:

import java.io.DataOutputStream;
import java.io.FileOutputStream;

public class Main
{

    public static void main( String[] args )
    {
        int x = 9;

        try
        {
            FileOutputStream fout = new FileOutputStream( "prueba.bin" );
            DataOutputStream salida = new DataOutputStream( fout );
            salida.writeInt( x );
            salida.close();
            System.out.println( "guardado" );
        }
        catch( Exception e )
        {
            System.out.println( e );
        }
    }
}

C++ Builder code:

unsigned char buffer[4];
ifstream file( "prueba.bin", ios::in | ios::binary );
if( file.is_open() )
{
    file.read( (char*)&buffer, sizeof( buffer ) );
    int num = buffer[ 0 ] | buffer[1] | buffer[ 2 ] | buffer[ 3 ];
    // another way to convert endianness:
    // int num = buffer[ 0 ] | ( buffer[1] << 8 ) | ( buffer[ 2 ] << 16 ) | ( buffer[ 3 ] << 24 );
}