CODEWITHSUNDEEP
javagenericsdiamond-operator
M Sach
M Sach

Reputation: 34424

Why Diamond operator was not missed from Right hand side in Java 7?

Java 7 has improved the diamond operator

In Java 6:

Map<String, String> myMap = new HashMap<String, String>();

In Java 7:

Map<String, String> myMap = new HashMap<>();

In Java 7 Types have been removed from the diamond operator on the right-hand side(RHS). My question why don't remove the complete diamond operator from RHS? I know it will throw the warning, but Java 7 could have removed the warning too.

  • Type safety: The expression of type HashMap needs unchecked conversion to conform to Map<String,String>
  • HashMap is a raw type. References to generic type HashMap<K,V> should be parameterized

The logic behind my thinking:
As we have already defined the map will have a string as a key and an object with Map<String, String> myMap on LHS. With this compiler has sufficient info. So why does it throw a warning if you miss the diamond operator altogether?

I am sure there must be a reason behind it but I am not getting it.

Upvotes: 2

Views: 718

Answers (1)

Cephalopod
Cephalopod

Reputation: 15145

The following code compiles and runs without error.

SoftReference<String> ref = new SoftReference(new Integer(1));
Object o = ref.get();
System.out.println(o); // prints "1"

A raw instance of SoftReference is created. "Raw" means that there is no generic type checking, which is required to allow to mix generics with pre-generics code.

By making the diamond operator implicit, you would break it.

Upvotes: 1

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