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Home/ Questions/Q 638679
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T20:45:32+00:00 2026-05-13T20:45:32+00:00

Code inside closures can refer to it variable. 8.times { println it } or

  • 0

Code inside closures can refer to it variable.

8.times { println it }

or

def mywith(Closure closure) {
   closure()
}

mywith { println it }

With this behavior in mind you can’t expect following code to print 0011

2.times {
   println it 

   mywith {
      println it
   }
}

And instead I have to write

2.times { i ->
   println i 

   mywith {
      println i
   }
}

My question is: why closures without parameters override it variable even if they don’t need it.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T20:45:32+00:00Added an answer on May 13, 2026 at 8:45 pm

    I think it has something to do with the formal Closure definition of Groovy:

    Closures may have 1…N arguments,
    which may be statically typed or
    untyped. The first parameter is
    available via an implicit untyped
    argument named it if no explicit
    arguments are named. If the caller
    does not specify any arguments, the
    first parameter (and, by extension,
    it) will be null.

    That means that a Groovy Closure will always have at least one argument, called it (if not specified otherwise) and it will be null if not given as a parameter.

    The second example uses the scope of the enclosing closure instead.

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