How does the Groovy in operator work?

Question

The Groovy \"in\" operator seems to mean different things in different cases. Sometimes x in y means y.contains(x) and sometimes it seems to call

Answer 1

It's actually all based on isCase. Groovy adds an isCase method to Collections that is based on the contains method. Any class with isCase can be used with in.

Answer 2

I did some experimentation and it looks like the in operator is based on the isCase method only as demonstrated by the following code

class MyList extends ArrayList {
    boolean isCase(Object val) {
        return val == 66
    }
}

def myList = new MyList()
myList << 55
55 in myList // Returns false but myList.contains(55) returns true     
66 in myList // Returns true but myList.contains(66) returns false

For the JDK collection classes I guess it just seems like the in operator is based on contains() because isCase() calls contains() for those classes.

Answer 3

in is the "Membership operator".

From the documentation for Groovy 3 (emphasis mine):

8.6. Membership operator

The membership operator (in) is equivalent to calling the isCase method. In the context of a List, it is equivalent to calling contains, like in the following example:

def list = ['Grace','Rob','Emmy']
assert ('Emmy' in list)           # (1)    

(1) equivalent to calling list.contains('Emmy') or list.isCase('Emmy')

So, Groovy always calls isCase, which in case of a List maps to contains.