How does java.util.Map's getOrDefault() work?

Question

I noticed that if I do map.getOrDefault("key1", new Object()), even if object is present for key1 in the map, new Object() is created. Though it is not returned by the method but it still creates it. For example,

public class Empl {     private int id;     private String name;      public Empl(String name) {         // TODO Auto-generated constructor stub         System.out.println(name);         this.name = name;     }     @Override     public String toString() {         // TODO Auto-generated method stub         return name+id;     } } 

running following,

Map<String, Empl> map = new HashMap<String, Empl>(); Empl imp = new Empl("timon"); map.put("1", imp); System.out.println(map.getOrDefault("1", new Empl("dumnba"))); 

gives this output:

timon dumnba timon0 

Shouldn't the default object be created only if it is not there in the map ? What is the reason if not ?

Answer 1

Shouldn't the default object be created only if it is not there in the map ?

How could that be the case? This call:

map.getOrDefault("1", new Empl("dumnba")) 

is equivalent to:

String arg0 = "1"; Empl arg1 = new Empl("dumnba"); map.getOrDefault(arg0, arg1); 

In other words, all arguments are evaluated before being passed to the method.

You could potentially use computeIfAbsent instead, but that will modify the map if the key was absent, which you may not want:

System.out.println(map.computeIfAbsent("1", k -> new Empl("dumnba"))); 

Answer 2

All the arguments to a function are evaluated before the function is executed. Java needs to evaluate new Empl("dumnba") so it can pass the result into getOrDefault. It can't know before getOrDefault is called that one of the arguments is not going to be required.

If you want to provide a default that is not computed unless needed, you can use computeIfAbsent. For this, you pass in a function, and that function is only executed if the default value is required.

map.computeIfAbsent("1", key -> new Empl("dumnba"))