函数.contains（）在预期的方式不能在Groovy中工作

Question

I am trying to check if number is member of list by using Groovy programming language.

I have this piece of code:

List<Long> list = [1, 2, 3]
Long number = 3

println(list.contains(number))​

Expected output is true , but the result I get is false .

Does anybody have an explanation?

Answer 1

Generic type parameters don't feature at runtime. Check this:

List<Long> list = [1, 2, 3]
list.each{println it.getClass()}

Which prints:

class java.lang.Integer
class java.lang.Integer
class java.lang.Integer

The true confusion is introduced by the bizarre behavior difference between .equals and == implementations:

Long.valueOf(3).equals(Integer.valueOf(3))
===> false
Long.valueOf(3) == Integer.valueOf(3)
===> true

List.contains seems to be using .equals , which checks the class of the parameter, thus explaining why forcing element types to Long resolves the problem.

So, in the midst of this uncertainty, I think the only sure thing is that Groovy's == execution performs the most intuitive and predictable comparison. So I'd change the check to:

boolean contains = list.grep{it == 3L} //sets value to true if matches at least 1

It helps when one doesn't have to be cognizant of data types linked to literals:

def ints = [1, 2, 3]
def longs = [1L, 2L, 3L]

boolean found1 = ints.grep{it == 3L}
boolean found2 = ints.grep{it == 3}
boolean found3 = longs.grep{it == 3L}
boolean found4 = longs.grep{it == 3}

println found1
println found2
println found3
println found4

Which works as anyone would want:

true
true
true
true

Answer 2

You should use the special literal values for longs: 1L instead of 1.

List<Long> list = [1L, 2L, 3L]
Long number = 3L

1 is an int. 1L is a long.