What is the most efficient way to create empty ListBuffer?

Question

What is the most efficient way to create empty ListBuffer ?

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

There are any pros and cons in difference?

Lylalyle

Answer 1

Order by efficient:

1. new mutable.ListBuffer[String]
2. mutable.ListBuffer.empty[String]
3. mutable.ListBuffer[String] ()

You can see the source code of ListBuffer & GenericCompanion

Engelbert

Answer 2

new mutable.ListBuffer[String] creates only one object (the list buffer itself) so it should be the most efficient way. mutable.ListBuffer[String] () and mutable.ListBuffer.empty[String] both create an instanceof scala.collection.mutable.AddingBuilder first, which is then asked for a new instance of ListBuffer.

Juliusjullundur

I searched for some source code. 'object Map' has def empty[A, B]: Map[A, B] = new HashMap[A, B] buf 'object ListBuffer' doesn't def empty. :( ListBuffer.empty looks have overhead as you say. Thank you – Lylalyle