In Java the types byte, short, int and long are all signed types, which
means that when a smaller type gets converted into a bigger type (short
to int for instance) the sign bit of the small type gets extended into
the larger type. This can happen when you make an explicite cast or when
the compiler implicitely promotes values to int or long before
evaluation in expressions. If your numbers really are unsigned you must
mask off the extended bits to "keep" them unsigned.

You could do

public static int getIntValue(short b) {
 return ((int)b)&0xFFFF;
}

Best regards,

I don't think it is the above portion of the code which is giving you
trouble. Calling your function with (short)0xE8 as the argument will
return 232 as the value  and not -24

I suspect the 0xE8 begins its life as a byte somewhere in your code.
(Reading it as a byte from the file ?)  - For instance :

byte b = (byte)0xE8;
int n= getIntValue(b);  // this will cause n to be -24 because byte is
an 8 bit signed entity

Incidentally, your getIntValue() function is largely unnecessary - As
Chris pointed out, a promotion to "int" is automatic for all
expressions involving integer types smaller than an int (short , char
or byte) - so a simple assignmet will suffice : i.e. the second
statement above does nothing more than

int n = b;

BK