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Comment on Java

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)

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karlthemailman@sh.itjust.works ⁨1⁩ ⁨year⁩ ago
So why not return a long or whatever the 64 bit int equivalent is?

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- whats_a_refoogee@sh.itjust.works ⁨1⁩ ⁨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).
  
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- korstmos@kbin.social ⁨1⁩ ⁨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 (?)
  
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parlaptie@feddit.de ⁨1⁩ ⁨year⁩ ago
But there’s really no point in flooring a double outside of the range where integers can be represented accurately, is there.

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affiliate@lemmy.world ⁨1⁩ ⁨year⁩ ago
what about using two ints

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- idunnololz@lemmy.world ⁨1⁩ ⁨year⁩ ago
  What about two int64_t
  
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  - affiliate@lemmy.world ⁨1⁩ ⁨year⁩ ago
    yeah that would be pretty effective. could also go to three just to be safe
    
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    hariette@artemis.camp ⁨1⁩ ⁨year⁩ ago
    Make it four, just to be even
    
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- korstmos@kbin.social ⁨1⁩ ⁨year⁩ ago
  A BigDecimal?
  
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