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javagenericstype-inference
Dheeru Mundluru
Dheeru Mundluru

Reputation: 383

Generic Method Type Argument Inference Issue

I am unable to understand why second invocation below gives compiler error. Why is it not able to infer Number as type argument as in the first invocation?

typeArgInference(new Integer[100], new ArrayList<Number>()); // Infers Number

typeArgInference(new Number[100], new ArrayList<Integer>()); // compiler error

<T> void typeArgInference(T[] a, Collection<T> c) {}

May be I am missing something here. If there is any rule in JLS on this behavior, please do include the link.

Upvotes: 0

Views: 76

Answers (1)

Andy Turner
Andy Turner

Reputation: 140319

Generics are invariant, arrays are covariant:

  • An Integer[] can act as a Number[] (because Integer is a subclass of Number)
  • An ArrayList<Integer> can't act as an ArrayList<Number> (because you can add e.g. a Double to the latter, but you mustn't add a Double into a list of Integers).

Of course, you can try to put a Double into a Number[], but it that would fail with ArrayStoreException if your Number[] is really an Integer[]. This would be a runtime failure.

Making arrays covariant was a fudge to have sort-of generic sort-of collections before the language supported generics. It was realized that this was a problem (because runtime failures suck), hence why generics were designed to be invariant.

Since you asked for JLS links:

Both of your examples would compile if you added an upper bound to the list type:

<T> void typeArgInference(T[] a, Collection<? extends T> c) {}

In both cases, Number is a type that would satisfy the type constraints:

  • Integer[] can act as a Number[]; and ArrayList<Number> is a Collection<Number>, which is a Collection<? extends Number>.
  • Number[] can act as a Number[] (obviously); and ArrayList<Integer> is a Collection<Integer>, which is a Collection<? extends Number>.

You just then could not add anything into the Collection within that method (other than literal null; or by breaking type safety via raw types).

Upvotes: 7

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