Can anyone please explain me below code
public <T extends Emp> void foo(ArrayList<T> list) {
list.add(list.remove(0)); // (cycle front element to the back)
}
I am trying to understand how Generics will apply on void return type. Note: i have changed the code now it is not giving any error.
I am trying to understand how generics will apply on void return type.
This is only a syntax. Generics do not apply to void return type, they apply to the method as a whole.
The name and restrictions on the generic parameter need to go somewhere in the text of your program. In a class declaration they follow the class name in angular brackets. In a generic method declaration they follow the accessibility designator public, and precedes the return type void.
One could easily imagine an alternative placement of generic types, such as
public void foo<T extends Emp>(ArrayList<T> list) // Imaginary syntax
or even
public void foo(ArrayList<T> list) <T extends Emp> // Imaginary syntax
but Java designers decided on the placement before the return type designator.
Generics won't be applied to void.
If you say that the type is <T extends Emp>, you are saying, that any subtype of Emp can be applied in place of T.
In your code, You can use <T> instead of <T extends Emp> as you aren't doing anything with Emp
public <T> void foo(ArrayList<T> list) {
list.add(list.remove(0)); // (cycle front element to the back)
}
Regarding how it'll work, the type will be provided by you when you use this method and at compile time, java will place required casts automatically. So, if you are using this:
ArrayList<String> list = new ArrayList<>();
// Add some items into list
foo(list);
in that case, your foo() method will find out that type <T> is String and so, will behave something like:
public void foo(ArrayList<String> list) {
list.add((String)(list.remove(0)));
}
Stringto anArrayList<T>whereTis definitely not a string.