Define bounded type parameter in method type parameter list or in method argument

user_5602950 0
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what's the difference between the following two method definitions? Which one is preferred if they are the same?

public static <T extends Comparable<? super T>>
void sort1(List<T> list) {}

public static <T>
void sort2(List<? extends Comparable<? super T>> list) {}

And if the methods have more then one parameter that uses the type parameter, what's the difference then?

public static <T extends Comparable<? super T>>
void add1(List<T> list, T element) {}

public static <T>
void add2(List<? extends Comparable<? super T>> list, T element) {}
generics java
1 Answers

These are not identical; for the type system, they are different. sort1 accepts a List of a type variable and sort2 accepts a List of a wildcard. Since you are doing a sort operation, that will modify the incoming parameter (the List); you might want to do this internally, for example:

public static <T extends Comparable<? super T>> void sort1(List<T> list) {
    list.set(0, list.get(0));
}

public static <T> void sort2(List<? extends Comparable<? super T>> list) {
    list.set(0, list.get(0)); // does not compile
}

only to find out that the second will not compile, because it uses a wildcard (there are ways to work-around this, though).

The same exact situation will arise in your second example, if you try to do:

public static <T> void add2(List<? extends Comparable<? super T>> list, T element) {
    list.add(element);
}

The general rule is that when you try to modify, you want to stick with type variables. If you can't do that, there are ways to work-around it, like wildcard capture, for example.

user_10593720
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