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Fix sequtils.delete bug with out of bounds indexes #12506

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Nov 29, 2019
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Fix sequtils.delete bug with out of bounds indexes #12506

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7 changes: 6 additions & 1 deletion lib/pure/collections/sequtils.nim
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Original file line number Diff line number Diff line change
Expand Up @@ -434,7 +434,9 @@ proc delete*[T](s: var seq[T]; first, last: Natural) =
var dest = @[1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1]
dest.delete(3, 8)
assert outcome == dest

doAssert first <= last
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Previously sending first > last resulted in a IndexError, which doesn't tell the user much about what went wrong. If sending first > last is supposed to work I can fix it instead.

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IMO, first greater than last should silently delete nothing. This how you define empty sequence with inclusive ranges.

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And first >= len should be IndexError

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And first >= len should be IndexError

I actually agree, but it would be inconsistent with how the last parameter is handled (out of bounds allowed) and how system.delete behaves. If first >= len results in IndexError then last >= len should also result in IndexError, which would be a breaking change.

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There is some Python influence here too, "oobasfd"[:10000] == 'oobasfd', Python is usually silent about index out of bounds for slices.

if first >= s.len:
return
var i = first
var j = min(len(s), last+1)
var newLen = len(s)-j+i
Expand Down Expand Up @@ -1156,6 +1158,9 @@ when isMainModule:
assert outcome == dest, """\
Deleting range 3-9 from [1,1,1,2,2,2,2,2,2,1,1,1,1,1]
is [1,1,1,1,1,1,1,1]"""
var x = @[1, 2, 3]
x.delete(100, 100)
assert x == @[1, 2, 3]

block: # insert tests
var dest = @[1, 1, 1, 1, 1, 1, 1, 1]
Expand Down