Java generics and type inference

Question

Type Inference

To illustrate this last point, in the following example, inference determines that the second argument being passed to the pick method is of type String:

static <T> T pick(T a1, T a2) { return a2; } Serializable s = pick("d", new ArrayList<String>());

Origirally I thought that the idea is that you could use any parameter in place of T as long as it ends up with String. Example ArrayList<ArrayList<String>>

But then I saw that the following also compiled fine:
Serializable s = pick("d", new ArrayList<Integer>());
It seems that T is inferred to be a Serializable and not a String?

So what is the meaning of the statement

inference determines that the second argument being passed to the pick method is of type String

@assylias:The most general form should be infered, right?

Cratylus
– Cratylus

2012-08-15 11:09:01 +00:00
Commented Aug 15, 2012 at 11:09 — Cratylus
– Cratylus, Commented Aug 15, 2012 at 11:09

assylias · Accepted Answer · 2012-08-15 11:13:01Z

2

In this case, the 3 types are Serializable, String, ArrayList<String>.

Serializable does not extend anything
String implements Serializable and other unrelated stuff
ArrayList<String> implements Serializable and other unrelated stuff

So the most specific type that applies to all 3 is Serializable.

If you replace the call with Serializable s = pick("d", new Object()); for example, it does not compile any longer because the most specific type is now Object and you can't cast Object to Serializable.

answered Aug 15, 2012 at 11:13

assylias

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5 Comments

Cratylus Over a year ago

1)Don't you mean ArrayList<E> implements Serializable?2) The most specific or the most generic type?

assylias Over a year ago

1) Yes ArrayList<Anything> implements Serializable 2) the most specific that applies to all 3. In your case, all three types are also Objects, but Serializable is more specific than Object so that's what is chosen.

Cratylus Over a year ago

If I change to: String s = pick("d", new ArrayList<String>()); it also does not compile.But this case is not clear why.It says: cannot convert from Serializable to String. How did it infer Serializable from this?All are Strings

assylias Over a year ago

Because Serializable is the most specific common supertype of both String and ArrayList.

martijno Over a year ago

Walk upwards the type hierarchy starting from Serializable, String, ArrayList<String> until you hit a common ancestor.

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