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javabinarybitset

Unexpected behaviour when doing bitwise operations halving bitset space

So I have for the longest time created "bitsets" in java.

I have always thought that if I use a long I have 64 bits to play with.

However I have now discovered it's not that simple...

And I suspect it has to with overflow and casting...

long x = 1<< 30;
return x+ "="+ Long.toBinaryString(x);

Renders:

1073741824=1000000000000000000000000000000

Expected.

But:

long x = 1<< 31;
return x+ "="+ Long.toBinaryString(x);

Renders:

-2147483648=1111111111111111111111111111111110000000000000000000000000000000

?????

I assume it has to with how the long equivalent value of the binary shift is being calculated and then rendered as a long. i.e. it is cast to a long and then calculated and this is an overflow I assume...

I expected

10000000000000000000000000000000

and the long value to be some negative number (can't figure out what it is).

There is definitely 64 bits in a long in java... as 

Long.toBinaryString(Long.MAX_VALUE);

Renders: 111111111111111111111111111111111111111111111111111111111111111 (64 1's).

How do I use the rest of the long space in my bitset?

Solution

  • This is because 1<< 31 is an int, equal to Integer.MIN_VALUE, which you then widen to a long.

    Make the first operand a long:

    1L << 31

    So this:

    long x = 1L << 31;
return x+ "="+ Long.toBinaryString(x);

    Returns:

    2147483648=10000000000000000000000000000000