What is the most efficient way to create emtpy ListBuffer ?
- 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 ?
views:
102answers:
2
+4
A:
Order by efficient:
new mutable.ListBuffer[String]mutable.ListBuffer.empty[String]mutable.ListBuffer[String] ()
You can see the source code of ListBuffer & GenericCompanion
Eastsun
2010-04-09 09:51:25
+5
A:
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.
Michel Krämer
2010-04-09 10:06:58
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
drypot
2010-04-09 23:48:52