What is the most efficient way to create emtpy ListBuffer ?

Posted by drypot on Stack Overflow See other posts from Stack Overflow or by drypot
Published on 2010-04-09T09:36:52Z Indexed on 2010/04/09 9:53 UTC
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scala

val l1 = new mutable.ListBuffer[String]
val l2 = mutable.ListBuffer[String] ()
val l3 = mutable.ListBuffer.empty[String]

There are any pros and cons in difference ?

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What is the most efficient way to create emtpy ListBuffer ? - Developer IT

What is the most efficient way to create emtpy ListBuffer ?

scala

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