• korstmos
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    501 year ago

    Doubles have a much higher max value than ints, so if the method were to convert all doubles to ints they would not work for double values above 2^31-1.

    (It would work, but any value over 2^31-1 passed to such a function would get clamped to 2^31-1)

      • @[email protected]
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        51 year ago

        To avoid a type conversion that might not be expected. Integer math in Java differs from floating point math.

        Math.floor(10.6) / Math.floor(4.6) = 2.5 (double)

        If floor returned a long, then

        Math.floor(10.6) / Math.floor(4.6) = 2 (long)

        If your entire code section is working with doubles, you might not like finding Math.floor() unexpectedly creating a condition for integer division and messing up your calculation. (Have fun debugging this if you’re not actively aware of this behavior).

      • korstmos
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        41 year ago

        Because even a long (64-bit int) is too small :)
        A long can hold 2^64-1 = 1.84E19
        A double can hold 1.79E308

        Double does some black magic with an exponent, and can hold absolutely massive numbers!

        Double also has some situations that it defines as “infinity”, a concept that does not exist in long as far as I know (?)

    • @[email protected]
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      41 year ago

      But there’s really no point in flooring a double outside of the range where integers can be represented accurately, is there.