Converting short, int and long to bytes.

Question

Hey,

This question is for the developers/everyone who knows about binary conversion on Lockergnome.net

I am currently having trouble with a custom network library which I plan to use for a lower data usage and therefore faster data transmission. It has to work on both C# and Java so therefore I will not be using any build in classes like BitConverter(C#).

My problem basically is that my numbers aren't being converted the way they should. When I pass a short with an value of 500 it returns 2 bytes of the values 244 and 0. This represents the short 244 not 500. With a value of 512 it returns 2 zero's

My current code in C#

short num = 500;

byte offset = 0xff;
byte b1 = (byte)(offset & num); // 244
byte b2 = (byte)((offset & num) >> 8); // 0, should be 1 since (1 * 256) + 244 = 500

//short has been converted to bytes (the wrong way)

short s = b1; // 244
s = (short)(s << b2); // 244 (didn't add anything for the 0 byte which is correct)

It doesn't matter which type of endian it uses as long as it transmits and receives the right data.

What am I doing wrong?

Answer 1

A right shift by 8 is equal to a division by 256, so (0xff & num) >> 8 is always 0. In fact, 0xff >> 8 is always 0. What makes you think otherwise?

Do the shifting before the & operation. Heck, you don't even need the & operation for this! The integers will wrap..

ushort val = 0xf00d;
byte x = (byte)val;
byte y = (byte)(val >> 8);

... or using division instead:

ushort val = 0xf00d;
byte x = (byte)val;
byte y = (byte)(val / 256);

Converting short, int and long to bytes.

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