Need help converting int to a 4 byte array

[kook04]

Hello all. I apologize in advance if this information is readily available, because I haven't been able to find a definitive answer to this question. I need to convert an int to a big endian 4 byte-array. I have the following code, which I'm fairly sure works, but it's the big endian part I'm not sure about. Can anyone confirm that this code snippet will do what I expect:

public static final byte[] intToByteArray(int value)
{
    return new byte[]
    { 
        (byte)(value & 0xff),
        (byte)(value >> 8 & 0xff),
        (byte)(value >> 16 & 0xff),
        (byte)(value >>> 24) };
    }
}

Thank you very much in advance for your help.

-Kennedy

[JosAH]

Hello all. I apologize in advance if this information is readily available, because I haven't been able to find a definitive answer to this question. I need to convert an int to a big endian 4 byte-array. I have the following code, which I'm fairly sure works, but it's the big endian part I'm not sure about. Can anyone confirm that this code snippet will do what I expect:

public static final byte[] intToByteArray(int value)
{
    return new byte[]
    { 
        (byte)(value & 0xff),
        (byte)(value >> 8 & 0xff),
        (byte)(value >> 16 & 0xff),
        (byte)(value >>> 24) };
    }
}

Thank you very much in advance for your help.

-Kennedy

Swap those four lines that initialize your array. Big endian means that the highest byte of the number comes first (opposite to little endian or 'intel endian').

kind regards,

Jos

[kook04]

Jos,

Are you saying that I should reverse the order of the elements in the array? It sounds like you are saying that the following would be big endian, is that true?

public static final byte[] intToByteArray(int value)
{
    return new byte[]
    { 
        (byte)(value >>> 24),
        (byte)(value >> 16 & 0xff),
        (byte)(value >> 8 & 0xff),
        (byte)(value & 0xff)
    };    
}

Thanks again.

-Kennedy

[JosAH]

Jos,

Are you saying that I should reverse the order of the elements in the array? It sounds like you are saying that the following would be big endian, is that true?

public static final byte[] intToByteArray(int value)
{
    return new byte[]
    { 
        (byte)(value >>> 24),
        (byte)(value >> 16 & 0xff),
        (byte)(value >> 8 & 0xff),
        (byte)(value & 0xff)
    };    
}

Yep, your first version stored the integer in little endian order and this version stores the int in big endian order in the array.

kind regards,

Jos

[kook04]

Jos,

Thank you again for your replies. If you have a moment, please take a look at the following methods. The intent is that these all output data in Big Endian format. However, based on your statements above, I think that they are actually providing Little Endian data.

public static int parseUnsignedShort(byte[] data, int offset)
{
    int returnValue = 0;

    returnValue = returnValue | (unsignedByteToInt(data[offset]));
    returnValue = returnValue | (unsignedByteToInt(data[offset + 1]) << 8);

    return StrictMath.abs(returnValue);
}

public static int parseUnsigned3ByteInt(byte[] data, int offset)
{
    int returnValue = 0;

    returnValue = returnValue | (unsignedByteToInt(data[offset + 2]) << 16);
    returnValue = returnValue | (unsignedByteToInt(data[offset + 1]) << 8);
    returnValue = returnValue | (unsignedByteToInt(data[offset + 0]) << 0);

    return StrictMath.abs(returnValue);
}

public static int parseUnsigned4ByteInt(byte[] data, int offset)
{
    int returnValue = 0;

    returnValue = returnValue | (unsignedByteToInt(data[offset + 3]) << 24);
    returnValue = returnValue | (unsignedByteToInt(data[offset + 2]) << 16);
    returnValue = returnValue | (unsignedByteToInt(data[offset + 1]) << 8);
    returnValue = returnValue | (unsignedByteToInt(data[offset + 0]) << 0);

    return StrictMath.abs(returnValue);
}

Thank you again, I really do appreciate your time.

-Kennedy

[JosAH]

Jos,

Thank you again for your replies. If you have a moment, please take a look at the following methods. The intent is that these all output data in Big Endian format. However, based on your statements above, I think that they are actually providing Little Endian data.

Yep, those methods all treat the byte arrays as low endian ints. Big endian ints as stored with their 'most important' byte at a lowest address, little endian ints are stored vice versa. You may want to look a ByteBuffers, they can fiddle diddle with the endianess of ints as well.

kind regards,

Jos