How to convert an int to a little endian byte array?

Question

I have this function in C# to convert a little endian byte array to an integer number:

int LE2INT(byte[] data) {   return (data[3] << 24) | (data[2] << 16) | (data[1] << 8) | data[0]; } 

Now I want to convert it back to little endian.. Something like

byte[] INT2LE(int data) {   // ... } 

Any idea?

Thanks.

Answer 1

The BitConverter class can be used for this, and of course, it can also be used on both little and big endian systems.

Of course, you'll have to keep track of the endianness of your data. For communications for instance, this would be defined in your protocol.

You can then use the BitConverter class to convert a data type into a byte array and vice versa, and then use the IsLittleEndian flag to see if you need to convert it on your system or not.

The IsLittleEndian flag will tell you the endianness of the system, so you can use it as follows:

This is from the MSDN page on the BitConverter class.

  int value = 12345678; //your value   //Your value in bytes... in your system's endianness (let's say: little endian)   byte[] bytes = BitConverter.GetBytes(value);   //Then, if we need big endian for our protocol for instance,   //Just check if you need to convert it or not:   if (BitConverter.IsLittleEndian)      Array.Reverse(bytes); //reverse it so we get big endian. 

You can find the full article here.

Hope this helps anyone coming here :)

Answer 2

Just reverse it, Note that this this code (like the other) works only on a little Endian machine. (edit - that was wrong, since this code returns LE by definition)

  byte[] INT2LE(int data)   {      byte[] b = new byte[4];      b[0] = (byte)data;      b[1] = (byte)(((uint)data >> 8) & 0xFF);      b[2] = (byte)(((uint)data >> 16) & 0xFF);      b[3] = (byte)(((uint)data >> 24) & 0xFF);      return b;   }