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Home/ Questions/Q 4545714
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Editorial Team
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Editorial Team
Asked: May 21, 20262026-05-21T15:51:07+00:00 2026-05-21T15:51:07+00:00

Is this madness, or is this Sparta? groovy:000> b = [1,2,3,4] ===> [1, 2,

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Is this madness, or is this Sparta?

groovy:000> b = [1,2,3,4]
===> [1, 2, 3, 4]
groovy:000> b.count { !it.equals(4) }
===> 0
groovy:000> b.count { !it == 4 }     
===> 0
groovy:000> b.count { it == 4 } 
===> 0
groovy:000> b.count { it == 1 }
===> 0
groovy:000> b[0]
===> 1
groovy:000> b.each { println it }
1
2
3
4
===> [1, 2, 3, 4]
groovy:000> print b.class
class java.util.ArrayList===> null
groovy:000> b.each { println it.class }
class java.lang.Integer
class java.lang.Integer
class java.lang.Integer
class java.lang.Integer
===> [1, 2, 3, 4]
groovy:000> 4.equals(b[3])
===> true
groovy:000> 

I’m running into a case of “surprised expectations” here. Groovy tells me that I have an ArrayList of Integer, and I expect that I should be able to do cute little searches like the above 3 queries all tersely and sweetly. But no.

  1. What is the idiomatic Groovy way of doing the above (count the number of elements where x != some element)
  2. Why doesn’t this work?
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-21T15:51:08+00:00Added an answer on May 21, 2026 at 3:51 pm

    be aware that the method signature

    public Number count(Closure closure)
    

    is supported since Groovy 1.8.0 (current production is 1.7.10) – see http://groovy.codehaus.org/groovy-jdk/java/util/Collection.html#count(groovy.lang.Closure)

    Before Groovy 1.8, the code above calls method ‘count(Object value)’, which counts the number of occurrences of the given value inside the collection. providing a closure instance as actual parameter ‘value’ leads to the results described above.

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