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7850 :: Is there a reliable way to round?

• xioren (orginal) [2021-04-24T18:34:52+02:00] view original

  I understand the warning about floating point decimals in the docs but other languages have suitable rounding functions. For example:

  nim:

  round(9.779999999999999, 2)
  Run

  give 9.779999999999999

  python:

  Run

  gives 9.78

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• Symb0lica (orginal) [2021-04-24T18:54:06+02:00] view original

  I'd suggest either

  
  import strutils
  echo formatFloat(9.779999999999999,format=ffDecimal,precision=2)
  Run

  or

  
  import strformat
  echo &"{9.779999999999999:0.2f}"
  Run

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• xioren (orginal) [2021-04-24T19:11:18+02:00] view original

  Thanks, will give those a try.

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• timothee (orginal) [2021-04-24T21:21:33+02:00] view original

  https://github.com/nim-lang/Nim/issues/13365 (ryu) is the proper way to fix this

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• juancarlospaco (orginal) [2021-04-25T00:08:37+02:00] view original

  other languages have suitable rounding

  https://0.30000000000000004.com

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• xioren (orginal) [2021-04-25T09:19:49+02:00] view original

  Run

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• reversem3 (orginal) [2021-04-27T17:53:08+02:00] view original

  import math echo round(9.779999999999999) output: echo round(9.779999999999999) 10.0

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• xioren (orginal) [2021-04-27T20:36:45+02:00] view original

  Try my example.

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• JohnAD (orginal) [2021-04-27T21:36:18+02:00] view original

  Not available yet, but the Decimals library is coming out in a future release of Nim based on the IEEE standard. It will support multiple rounding methods including standard bankers rounding (aka rounding to even).

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• xigoi (orginal) [2021-04-28T08:21:09+02:00] view original

  The number 9.78 can't be represented exactly as a floating-point value. Python is just implicitly rounding the resulting value.

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• reversem3 (orginal) [2021-04-28T20:33:07+02:00] view original

  Thats python syntax , I'm showing how to in nim by using the math module.

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• jxy (orginal) [2021-04-30T23:14:47+02:00] view original

  Certain language just likes to go that extra mile to lie to you.

  
  >>> round(9.779999999999999, 2)
  9.78
  >>> (9.78).hex()
  '0x1.38f5c28f5c28fp+3'
  >>> round(9.779999999999999, 2).hex()
  '0x1.38f5c28f5c28fp+3'
  >>> (9.779999999999999).hex()
  '0x1.38f5c28f5c28fp+3'
  Run

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• timothee (orginal) [2021-06-21T10:18:00+02:00] view original

  this is now fixed now that we have dragonbox

  when true: # D20210513T184615
    from math import round
    let a = round(9.779999999999999, 2)
    doAssert a == 9.78
    echo a # before dragonbox: 9.779999999999999, after dragonbox: 9.78, like in python3
  Run

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