CODEWITHSUNDEEP
javagenerics
ShaggyInjun
ShaggyInjun

Reputation: 2963

Java generics clarification of a code snippet

I am trying to understand this piece of Code. It is from Oracle Generics Page.

I am seeing two return types here <T extends Comparable<T>> and int. Am I reading this right ? If so how can a method have two return types ?

public static <T extends Comparable<T>> int countGreaterThan(T[] anArray, T elem) {
    int count = 0;
    for (T e : anArray)
    if (e.compareTo(elem) > 0)
        ++count;
    return count;
}

Upvotes: 1

Views: 96

Answers (2)

Guillaume
Guillaume

Reputation: 1879

The method has only one return type: int.

<T extends Comparable<T>> is just to declare the type of T.

If you do something like this:

public static int calculate(T param) {
  ....
}

You would have a compilation error as T is undefined. T is a generic type so you need to specify it:

public static <T> int calculate(T param) {
  ....
}

Upvotes: 1

Miserable Variable
Miserable Variable

Reputation: 28752

No, the return type is int

T extends Comparable<T> is type parameter, and used in the parameters.

Upvotes: 3

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