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Champagne Floats

Suppose you've got a string that contains a floating point number: "3.5" or "2.5E-3". Your goal is to get a float from that number. There are a few different techniques to try...

If you're familiar with parseDouble(), you could use that and then cast:
    String s = ...;
    float a = (float) Double.parseDouble(s);

Or you could save some bits and parse directly using parseFloat():
    String s = ...;
    float b = Float.parseFloat(s);


Do the two mechanisms yield the same result?


Ponder the problem, and when you're ready you can read my analysis in the comments...

# posted by Jesse Wilson on Wednesday, October 28, 2009

Neither floats nor doubles have infinite precision. For some number n1 exactly expressible as a float, there's a next float n2. Between the two exist real numbers that have no perfect representation as a float.

But many of these real numbers between n1 and n2 do have an exact representation as a double.

To create an analogy, suppose that floats can represent only whole numbers: 1.0, 2.0, 3.0. Similarly, doubles can represent both whole numbers and half numbers: 1.0, 1.5, 2.0, 2.5, 3.0.

Given the input string "1.4":
A: the nearest float is 1.0
B: the nearest double is 1.5
C: the float nearest to B is 2.0

And so we can intuit that the two mechanisms will not yield the same result for values like "1.4" that would get rounded twice. For doubles and floats, "2.1019476964872256E-45" is such a value.

# posted by Blogger swankjesse on October 28, 2009 1:17 AM

Could you please remove the tons of trojans/viruses on publicobject.com. Chrome warns about it. I almost lost my laptop getting the "Antivirus System Pro" when downloading the glazed lists impl...

# posted by Blogger MrCoder on October 29, 2009 5:25 AM

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