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13406 :: Idiomatic way to check if value is in subrange?

• nrk (orginal) [2025-09-17T22:08:30+02:00] view original

  See title. I've tried:

  
  type
    A = enum
      a, b, c
    
    Subrange = range[b..c]
  
  #echo A.b in Subrange # doesn't compile
  
  echo A.b in Subrange.low..Subrange.high # looks fine, except...
  echo A.a in Subrange.low..Subrange.high # RangeDefect at runtime :(
  
  echo A.a in Subrange.low.A..Subrange.high.A # finally works, but...really?
  Run

  Am I missing something?

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• Zoom (orginal) [2025-09-17T22:53:58+02:00] view original

  
  type
    A = enum
      X, Y, Z
    
    B = Y..Z
  
  doAssert Y isnot B
  
  const Range = B.low..B.high
  
  doAssert Y in Range
  
  let b = B(Y)
  
  doAssert b is B
  doAssert b in Range
  doAssert b is A
  
  type SR = range[B.low..B.high]
  
  doAssert Y isnot SR
  
  let yr = SR.low
  
  doAssert yr is SR
  
  doAssert yr is B
  # ...but requires conversion for direct comparison
  doAssert yr.B == B.low
  Run

  in check for ranges, is checks for types.

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• 3-2-1 (orginal) [2025-09-17T23:24:31+02:00] view original

  I think he's more interested in knowing why is the behavior different from the expected 0 notin 1..2

  
  type
    A = enum
      a, b, c
    Subrange = range[b..c]
  
  assert 0 notin 1..2
  
  try:
    echo "A.a in Subrange.low..Subrange.high = " & $(A.a in Subrange.low..Subrange.high)
  except RangeDefect:
    echo "raises RangeDefect"
  Run

  This outputs,

  
  A.a in Subrange.low..Subrange.high = false
  raises RangeDefect
  Run

  My question is why it throws after evaluating correctly.

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