Generic syntax symbol <~> [duplicate] - java

This question already has answers here:
What does a tilde in angle brackets mean when creating a Java generic class?
(4 answers)
Closed 9 years ago.
Working with IDE like NetBeans or IDEA, i've seen that they are converting generic types into this symbol:
private final List<String> ar = new ArrayList<~>();
But using this in a simple editor results in throwing an error. BTW Eclipse also doesn't like it. Is it somehow connected with type erasure mechanism?

You will see that in IntelliJ its a different colour!
This is because it has used code folding to hide the code, but you can't write it this way in Java 6.
In Java 7 you can write
private final List<String> ar = new ArrayList<>();

Related

Run Time Polymorphism in Java Collection framework [duplicate]

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What does it mean to "program to an interface"?
(33 answers)
Java - declaring from Interface type instead of Class
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Closed 11 months ago.
Set set= new TreeSet<>();
vs
TreeSet set= new TreeSet<>();
I know we are using the concept of run time polymorphism here, but what is the use of writing this? Is there any advantage to writing one thing over another? both are giving me the correct results. I am learning java right now, help would be great.

Java HashMap declaration [duplicate]

This question already has answers here:
What is the point of the diamond operator (<>) in Java?
(7 answers)
Closed 2 years ago.
Hi Guys I have a question about HashMap declaration in Java. I'd like to know what is the difference between:
HashMap<String,Integer> map = new HashMap<>();
and
HashMap<String,Integer> map = new HashMap<String,Integer>();
Most coding websites teach the second one. However, the first one works perfectly fine. Is there a reason we should use second one?
There is technically none, however since Java 1.7 it's a convention to use this <> diamond operator, it reduces the text size.
This is called Type Inference.
Compilers from releases prior to Java SE 7 are able to infer the actual type parameters of generic constructors, similar to generic methods. However, compilers in Java SE 7 and later can infer the actual type parameters of the generic class being instantiated if you use the diamond (<>).
new HashMap<String,Integer>() - The type parameter is mandatory before Java 7
new HashMap<>() - The type parameter is inferred as of Java 7
It's recommended to use the "diamond operator" (<>) due to the brevity. Effectively, there is no difference.
There is no difference between the two of those since form the Java 7. Previously before Java 7 you must mention the Generic type parameters like in your second case in both sides.But to make it easier so that developer can write less code after Java 7 only you need to define it once in the left side.
Before Java Version 7 you need to code like below.
HashMap<String,Integer> map = new HashMap<String,Integer>();//For java versions below 1.7
After Java version 7
HashMap<String,Integer> map = new HashMap<>();//For java versions above 1.7

What is purpose of type on the constructor side of Java collection creation? [duplicate]

This question already has answers here:
Is there a difference between explicitly putting the type into the diamond operator vs letting java figure it out?
(4 answers)
Closed 5 years ago.
What is purpose of re-specifying the type on the right / constructor side of a Java Collection declaration + instantiation?
For example, how is
List<MyClass> a = new ArrayList<MyClass>();
different / better / worse than
List<MyClass> a = new ArrayList<>();
My understanding is the the second form became legal in Java 7, but I still see a lot of examples in Java 7 and 8 where the first form is used.
It is not worse or different. It is the same.
A new and easy way introduced by compiler to reduce human effort of defining of type.

What is the difference between "ArrayList<String> = new ArrayList<>()" and "ArrayList<String> = new ArrayList<String>()"? [duplicate]

This question already has answers here:
What is the diamond operator in Java?
(2 answers)
Closed 7 years ago.
just like the title said, in Java, is there any difference between the two statements?
Usually I can see the two statements both. So I suppose there may be no difference. But could you help to confirm and explain why they are identical?
ArrayList<String> al = new ArrayList<String>();
ArrayList<String> al = new ArrayList<>();
Thanks!
They are identical. new ArrayList<String>(); was required prior to JDK 7. In JDK 7 they introduced the 'diamond operator', where the type is inferred.
So, if you are using JDK 7+ you can simply use new Arraylist<>()
You can see this in the Java Language Specification under Class Instance Creation Expressions, or search for 'Diamond Operator'
This is a Diamond Operator.
The Diamond Operator reduces some of Java's verbosity surrounding generics by having the compiler infer parameter types for constructors of generic classes.
http://www.javaworld.com/article/2074080/core-java/jdk-7--the-diamond-operator.html
In java 7 and later there is no difference. The latter is considered more concise.
The so called diamond operator <> is helping here for the type to be inferred.

Error Casting in Java [duplicate]

This question already has answers here:
Converting a generic argument to an int in java, provided that it is a number
(4 answers)
Closed 8 years ago.
I have the next problem:
I am making a simple cast:
TargetBot2Params params = (TargetBot2Params)bot.getParams();
But I get the next error:
Inconvertible types
Requiered: TargetBot2Params
Found: UT2004BotParameters
And I dont know why occurs this, because I think that it doesnt do the cast.
Anyone knows why occurs this?
Thanks for your time.
The error message says it all.
For the casting to be successful, UT2004BotParameters has to either be a TargetBot2Params (i.e., extends TargetBot2Params) or, if TargetBot2Params is an interface, implement it.

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