Get the first element of a list idiomatically in Groovy

Question

Let the code speak first

def bars = foo.listBars()
def firstBar = bars ? bars.first() : null
def firstBarBetter = foo.listBars()?.getAt(0)

Answer 1

Not sure using find is most elegant or idiomatic, but it is concise and wont throw an IndexOutOfBoundsException.

def foo 

foo = ['bar', 'baz']
assert "bar" == foo?.find { true }

foo = []
assert null == foo?.find { true }

foo = null
assert null == foo?.find { true }  

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--Update Groovy 1.8.1
you can simply use foo?.find() without the closure. It will return the first Groovy Truth element in the list or null if foo is null or the list is empty.