Parsing binary file- unsigned longs

Question

Hello everyone.
I'm currently trying to write a quick parser for a binary file format (ASF). However, java's handling of bit shifting and lack of unsigned types is driving me completely mad.
How can I represent an unsigned long?
How can I set the bits for it anyway? It seems that the shift operator can't shift more than an int's worth (it just loops around and starts adding to the other bits- weird).
Thanks in advance.
Simon

Answer 1

ejp wrote:
But why in the world does the following code also do nothing?
long x = 45 << 32;
Try long x = 45L << 32;
The answer appears to be that java uses ints to represent all constants, but this presents some serious problems- namely- how on earth do you get a number bigger than an int into a long?With 'L'. Same as in C or C++. In Visual Basic it's '&'. There's generally something.Where did that come from? Why have I never seen anything like that before?
If I do long x = 0x7FFFFFFF; all is well, but if I do long x = 0x80000000; I get FFFFFFFF80000000, which isn't what I asked for.Actually it is exactly what you asked for. You've overlooked several facts: (i) Java longs are 64 bits; (ii) you specified a negative 32-bit constant; (iii) the rules of every programming language I can think of provide for sign-extension when going from a shorter to a longer type.Right. It makes sense now that I know how to specify long constants.
As someone pointed out signed/unsigned is actually the same, so long as you interpret it correctly. So to avoid the total stupidity of using twice as much memory for everything I've decided that I am actually going to use the correct types.They're not the correct types. As I pointed out in an earlier post, your 'unsigned longs' must be coming from C or C++ where they are generally 32 bits, so using a 64-bit Java long is incorrect.Where they came from doesn't matter. The spec doesn't say it's a "long"- it says that this particular value is 64bit, and that all values are unsigned. So I do need a Java long.
WHY IN THE WORLD IS JAVA "INTELLIGENT" WHEN DOING THINGS BITWISE? WHICH BRAIN DEAD IDIOT THOUGHT THAT UP? That is broken and is asking for trouble.It is you who is asking for trouble here. The rules of Java are consistent and in conformity with the practice in other programming languages. You've just overlooked several relevant facts.I think I've worked out where I was going wrong here. When doing something like
int i;
long x;
x = x | i;The i is converted to a long before the bitwise operation, so it's not the bitwise operation that's the problem, it's the conversion between int and long?
It's not Java whose stupidity is the issue here ;-)That wouldn't surprise me.
Thanks.