groovy

closures do not handle 'it' correctly

Details

  • Type: Bug Bug
  • Status: Closed Closed
  • Priority: Minor Minor
  • Resolution: Not A Bug
  • Affects Version/s: 1.8.5
  • Fix Version/s: None
  • Component/s: groovy-jdk
  • Labels:
  • Environment:
    Windows command line (to reduce dependencies)
  • Number of attachments :
    0

Description

Running the following script shows 'it' as being null:

dec = { it ->
def x = it
return { println "$

{x--}

(was $it)" }
}

def f = dec(2)
f(); f();

Output is:
2 (was null)
1 (was null)

If the 'it' references are changed to 'val', then it works fine - i.e. says '(was 2)'

This also fails when no parameter is used (i.e. defaulting to 'it')

Activity

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Andrew Stratton added a comment -

Please see comments if necessary.

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Andrew Stratton added a comment - Please see comments if necessary.
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Andrew Stratton added a comment -

Hi Paul
Yes - I get it now - its the second closure that has 'it' nulled out.

e.g.
f1 = { val ->
def x = val-1
return

{ println "$val -> $x" }

}
(f1(2))()
===
2 -> 1
This works, because val exists in the scope outside of the second closure call.

f2 = { it ->
def x = it-1
return

{ println "$it -> $x" }

}
(f2(2))()
(f2(2))(3)
===
null -> 1
3 -> 1

This shows what is really happening with the original example - 'it' referred to the parameterless call to second closure.

One last example to confirm:
f3 = { it ->
def x = it-1
return

{ y -> println "$it -> $x" }

}
(f3(2))()
(f3(2))(3)
===
2 -> 1
2 -> 1

Yep - the 'it' parameter is initialized to the passed parameter - or null if none is provided - but when aparameter name is used, 'it' is left alone.

Cheers
Andy

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Andrew Stratton added a comment - Hi Paul Yes - I get it now - its the second closure that has 'it' nulled out. e.g. f1 = { val -> def x = val-1 return { println "$val -> $x" } } (f1(2))() === 2 -> 1 This works, because val exists in the scope outside of the second closure call. f2 = { it -> def x = it-1 return { println "$it -> $x" } } (f2(2))() (f2(2))(3) === null -> 1 3 -> 1 This shows what is really happening with the original example - 'it' referred to the parameterless call to second closure. One last example to confirm: f3 = { it -> def x = it-1 return { y -> println "$it -> $x" } } (f3(2))() (f3(2))(3) === 2 -> 1 2 -> 1 Yep - the 'it' parameter is initialized to the passed parameter - or null if none is provided - but when aparameter name is used, 'it' is left alone. Cheers Andy
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Paul King added a comment -

Is this not what is expected? You are returning a closure - you have not supplied any args to that closure, hence "it" for the closure is null (even though the outer closure will have a different "it" value).

Show
Paul King added a comment - Is this not what is expected? You are returning a closure - you have not supplied any args to that closure, hence "it" for the closure is null (even though the outer closure will have a different "it" value).

People

  • Assignee:
    Unassigned
    Reporter:
    Andrew Stratton
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Dates

  • Created:
    Updated:
    Resolved: