Further, why would you use something designed to store floating point decimal numbers when you are computing what is clearly an integer value?

If you need something more accurate, look at something like the BitInteger class.

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

fred rosenberger wrote:If you need something more accurate, look at something like the BitInteger class.

I think that's

`BigInteger`, Fred.

@Shamsudeen - More specifically, something like:

`private static final BigInteger TWO = BigInteger.valueOf(2L);`(I don't know why BigInteger doesn't define this constant itself)

and then elsewhere in your code:

`TWO.pow(1000);`

Winston

"Leadership is nature's way of removing morons from the productive flow" - Dogbert

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All the primitives have limits of range, and the floating-point types have limits to their precision, too. The limits are clearly stated in the Java Language Specification and Java Tutorials, and you can find more by Googling for IEEE754. 2 to the 1000 is actually inside the range of aShamsudeen Akanbi wrote: . . . With "double x = Math.pow(2,1000)". The result is deviating, i mean its rounding it off at a particular limit . . .

`double`, so you won’t get infinitiy, but it will tell you it is only precise to 15 or 16 Significant figures. Floating-point arithmetic is designed for engineers to use.

When I was an undergraduate:

Floating-point arithmetic is intended for people who don’t mind such imprecision. You have already been given the correct solution, twice, with and without spelling errorsWhat’t an engineer?

He’s somebody you ask “What’s 2×2?”, and he gets his slide‑rule and says, “ . . . two by two . . . three point nine nine . . . that’s four, near as makes no difference.”

Campbell Ritchie wrote:When I was an undergraduate: What's an engineer?...

Nice one.

Winston

"Leadership is nature's way of removing morons from the productive flow" - Dogbert

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Winston Gutkowski wrote:TWO.pow(1000);Winston

BTW: If you fancy it (and you have the memory space and - possibly more importantly - time )

try:

BigInteger FOUR = new BigInteger.valueOf(4L);

`System.out.println(FOUR.pow(1073741824).bitLength());`

I've never got it to work; probably because my heap memory is too small (possibilities run to anywhere around 3Gb, depending on efficiency) and I can't be bothered to fix it; but I suspect very strongly that it will display a negative number..SILENTLY. And this after repeated queries about it to Oracle and Sun.

Winston

"Leadership is nature's way of removing morons from the productive flow" - Dogbert

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Campbell Ritchie wrote:You already know its bit length. I can tell you it free. 2147483649.

And

`BigInteger.bitLength()`returns

*what*?

Winston

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Campbell Ritchie wrote:2147483649. Presumably rounded with the ROUND_PEG_SQUARE_HOLE algorithm, to the nearestint.

. . . And how long does it take to run? (Or crash?)

Actually, nanoseconds; unless the number is negative and an exact power of 2; so, on average...

My objection is theoretical:

`bitLength()`returns an

`int`, which has a limit of 23^31 -1. The magnitude of a BigInteger is held in an

`int[]`, which can hold up to ≈2^36 bits. I've just never been able to prove that it does what I think it will.

Winston

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How can a number be negative and an exact power of 2?Winston Gutkowski wrote: . . . unless the number is negative and an exact power of 2; so, on average... . . .

I did work out that something would go wrong, so I said it would use the ROUND_PEG_SQUARE_HOLE algorithm. I have had the d*mn thing running for several hours with no sign of any output yet.

Campbell Ritchie wrote:How can a number be negative and an exact power of 2?Winston Gutkowski wrote: . . . unless the number is negative and an exact power of 2; so, on average... . . .

I guess I should have said -(2^n).

BigIntegers hold their values in sign/magnitude form, and

`bitLength()`is defined as returning:

*"the number of bits in the*

__minimal__two's-complement representation of this BigInteger, excluding [the] sign bit."I take that to mean "the number of bits from the first one in the 2's-C form that

*differs*from the sign bit"; otherwise the check doesn't make much sense.

Personally, I think they could've saved themselves a lot of bother by simply returning the bit length of the absolute value, but maybe they had their reasons.

BTW: How did your attempt work out?

Winston

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