Java Generic Method Question

Question

Consider this code:

public <T> List<T> meth(List<?> type)
{
   System.out.println(type); // 1
   return new ArrayList<String>(); // 2
}

It does not compile at line 2, saying that List is required.

Now, if it's changed to:

public <T> List<?> meth(List<T> type)
{
   System.out.println(type); // 1
   return new ArrayList<String>(); // 2
}

It does compile. Why? I thought the difference between declaring a generic type with T and using the wildcard was that, when using the wildcard, one cannot add new elements to a collection. Why would <?> allow a subtype of List to be returned? I'm missing something here, what's the explicit rule and how it's being applied?

What is your original goal? What do you want to accomplish with your method? — akarnokd
– akarnokd, Commented Jul 19, 2009 at 14:09
Nothing really, this isn't part of a project, I just wanna understand java generic methods. I' reading this right now java.sun.com/j2se/1.5/pdf/generics-tutorial.pdf — andandandand
– andandandand, Commented Jul 19, 2009 at 14:11

notnoop · Accepted Answer · 2009-07-19 15:34:12Z

4

The difference is in the return type declaration. List<String> is not a subtype of List<T>, but it is a subtype of List<?>.

List<?> makes no assumptions regarding its type variable, so the following statements are valid:

List<?> l0 = new ArrayList<String>();
List<?> l1 = new ArrayList<Object>();
List<? extends Number> ltemp = null;
List<?> l2 = ltemp;

List<T> assumes that the type argument will be resolved in the context of client (e.g. type use), when you declared it as List<String> or List<Object>. Within the method body, you cannot make any assumptions about it either.

edited Jul 19, 2009 at 15:34

answered Jul 19, 2009 at 14:10

notnoop

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1 Comment

Tom Hawtin - tackline Over a year ago

I don't think the last compiles.

Mehrdad Afshari · Accepted Answer · 2009-07-19 14:14:57Z

1

In the first case, T is not necessarily a superclass of String. If you choose a T like Integer and call the method, it'll fail; so it won't compile. However, the second will compile as surely, any ArrayList<String> is a valid List of something.

edited Jul 19, 2009 at 14:14

answered Jul 19, 2009 at 14:09

Mehrdad Afshari

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Comments

Valentin Rocher · Accepted Answer · 2009-07-20 07:29:35Z

1

As said earlier, String isn't a subtype of T, so it's why it does not work. However, this code works :

public <T> List<T> meth(List<?> type)
{
   System.out.println(type); // 1
   return new ArrayList<T>(); // 2
}

and is more in the idea of what you want, I think.

answered Jul 20, 2009 at 7:29

Valentin Rocher

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