Read two bytes into an integer?

Question

I have a byte[] that I've read from a file, and I want to get an int from two bytes in it. Here's an example:

byte[] bytes = new byte[] {(byte)0x00, (byte)0x2F, (byte)0x01, (byte)0x10, (byte)0x6F}; int value = bytes.getInt(2,4); //This method doesn't exist 

This should make value equal to 0x0110, or 272 in decimal. But obviously, byte[].getInt() doesn't exist. How can I accomplish this task?

The above array is just an example. Actual values are unknown to me.

Answer 1

You should just opt for the simple:

int val = ((bytes[2] & 0xff) << 8) | (bytes[3] & 0xff); 

You could even write your own helper function getBytesAsWord (byte[] bytes, int start) to give you the functionality if you didn't want the calculations peppering your code but I think that would probably be overkill.

Answer 2

Try:

public static int getInt(byte[] arr, int off) {   return arr[off]<<8 &0xFF00 | arr[off+1]&0xFF; } // end of getInt 

Your question didn't indicate what the two args (2,4) meant. 2 and 4 don't make sense in your example as indices in the array to find ox01 and 0x10, I guessed you wanted to take two consecutive element, a common thing to do, so I used off and off+1 in my method.

You can't extend the byte[] class in java, so you can't have a method bytes.getInt, so I made a static method that uses the byte[] as the first arg.

The 'trick' to the method is that you bytes are 8 bit signed integers and values over 0x80 are negative and would be sign extended (ie 0xFFFFFF80 when used as an int). That is why the '&0xFF' masking is needed. the '<<8' shifts the more significant byte 8 bits left. The '|' combines the two values -- just as '+' would. The order of the operators is important because << has highest precedence, followed by & followed by | -- thus no parentheses are needed.