Initializing arraylist without object type - JAVA [duplicate]

问题

This question is not about why we initialize a list as interface over implementation e.g.

List<myObject> obj = new ArrayList<myObject>();

The question is what is the difference between the following two and why do they (apparently) work the same way?

//list and arraylist both have a type
List<myObject> obj = new ArrayList<myObject>();

//arraylist does not have a type
List<myObject> obj = new ArrayList<>();

回答1:

Both pieces of code are equivalent and create ArrayLists with a type (myObject in your example):

List<myObject> obj = new ArrayList<myObject>();

List<myObject> obj = new ArrayList<>();

However the second example uses the diamond operator (<>) introduced in Java 7. It adds type inference and reduces the verbosity in the assignments.

See the following quote from the documentation:

You can replace the type arguments required to invoke the constructor of a generic class with an empty set of type parameters (<>) as long as the compiler can infer the type arguments from the context. This pair of angle brackets is informally called the diamond.

For example, consider the following variable declaration:
Map<String, List<String>> myMap = new HashMap<String, List<String>>();
In Java SE 7, you can substitute the parameterized type of the constructor with an empty set of type parameters (<>):
Map<String, List<String>> myMap = new HashMap<>();

回答2:

In your second example, Java assumes the type is myObject. So there is still a type.

来源：https://stackoverflow.com/questions/47413609/initializing-arraylist-without-object-type-java

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java

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arraylist