CODEWITHSUNDEEP
javac++
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Reputation: 1519

convert c++ method logic in java by using right data type

I have few fields in C++ defined and using those fields we are making a new value. I need to do similar things in Java by following C++ method.

Below are C++ methods:

typedef uint64_t    ClientData;

// convert all four fields into one variable "client_data"
void pack(uint8_t text, 
          uint16_t client_id, 
          uint8_t proc_id, 
          uint32_t pod_count,
          ClientData& client_data)
{
    client_data = (uint64_t(text) << 56)
                   + (uint64_t(client_id) << 40)
                   + (uint64_t(proc_id) << 32)
                   + pod_count;
}

// now retrieve individual fields value from converted "client_data" variable
void unpack(ClientData client_data,
            uint8_t& text, 
            uint16_t& client_id, 
            uint8_t& proc_id, 
            uint32_t& pod_count)
{
    pod_count = client_data & 0xffffffff;
    client_data = client_data >> 32;
    proc_id = client_data & 0xff;
    client_data = client_data >> 8;
    client_id = client_data & 0xffff;
    client_data = client_data >> 16;
    text = client_data & 0xff;
}

Now this is what I am doing in Java:

public final class ClientData {
  private final byte text;
  private final short clientId;
  private final byte procId;
  private final int podCount;

  // this is my pack method corresponding to c++ method above
  public long pack() {
    return (((long) text) << 56) | (((long) clientId) << 40) | (((long) procId) << 32)
        | ((long) podCount);
  }  

  // this is my unpack in Java.
  public unpack(long packedValue) {
    this.text = (byte) ((packedValue >>> 56) & ((1 << 8) - 1));
    this.clientId = (short) ((packedValue >>> 40) & ((1 << 16) - 1));
    this.procId = (byte) ((packedValue >>> 32) & ((1 << 8) - 1));
    this.podCount = (int) ((packedValue) & ((1L << 32) - 1));
  }
}

Now my question is - Whether the logic that I have in pack and unpack method in Java is correct as per above C++ code? And pod_count in C++ is defined as uint32_t but in my above Java code I have it as int.

Upvotes: 0

Views: 238

Answers (1)

damjad
damjad

Reputation: 1290

By default, Java has signed numbers but you can use them as unsigned ones. Java Primitive data types.

  • uint8_t: byte is a 8-bit signed integer. If you want to add a sign-bit, you have to use short.
  • uint16_t: short is a 16-bit signed integer. If you want to add a sign-bit, you have to use int.
  • uint32_t: int is a 32-bit signed integer. But, you can use it as an unsigned integer. For that, you might wanna look into the unsigned comparison methods etc.
  • uint64_t: long is a 64-bit signed integer. But, you can use it as an unsigned integer.

I think it does not matter if you are using signed or unsigned of data types are using with the bitwise operators. The only problem you will get it is while printing those numbers or converting them to strings. For that, you can use either toUnsignedInteger method of Byte and Short or toUnsignedLong of Integer.

For the logic part: You can implement the bitwise logic as in C++ but with type casts. Please refer to the documentation for more information.

public class ClientData
{
    private int pod_count;
    private byte proc_id;
    private short client_id;
    private byte text;

    public static void main (String[] args)
    {
        new ClientData().unpack(Long.decode("0x32112321321fdffa"));
    }
    void unpack(long client_data)
    {
        this.pod_count = (int)(client_data & 0xffffffff);
        client_data = client_data >> 32;
        this.proc_id = (byte) (client_data & 0xff);
        client_data = client_data >> 8;
        this.client_id = (short) (client_data & 0xffff);
        client_data = client_data >> 16;
        this.text = (byte) (client_data & 0xff);
    }
}

Upvotes: 1

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