JDK-6488666 : inference: lub of typevariables does not yield expected results
  • Type: Enhancement
  • Component: specification
  • Sub-Component: language
  • Affected Version: 5.0,6
  • Priority: P4
  • Status: Open
  • Resolution: Unresolved
  • OS: generic
  • CPU: generic
  • Submitted: 2006-11-01
  • Updated: 2015-04-17
Related Reports
Duplicate :  
JDK-6557665 - Type variables are erased during lub computation
Relates :  
JDK-6357966 - parameters containing nested generic types avoid recursive calls
Relates :  
JDK-5042462 - Problem with least upper bound (lub) and type variables
Relates :  
JDK-6569277 - inference: uninferred type args should follow their upper bounds
Description
Consider the example in 5042462.

We have the following type variables:

<T, U extends T, V extends T>

The JLS (p463) defines lub in terms of the EST (erased supertype set).

EST(U) = {Object} and
EST(V) = {Object}.

Then EC = { Object } = MEC.

So lub(U,V) = Object.

It should be T.
Comments
EVALUATION I am inclined to think we should pass the bounds of every type argument to the inference process and use them in the lub calculation. Consider the example in 6557661, slightly modified below: <T> Map<T,T> foo(T t1, T t2) { return null; } <U extends Object, V extends U> Map<U,U> bar(U u, V v) { Map<Object,Object> map1 = foo(u, v); // Allowed by javac 1.6 Map<U,U> map2 = foo(u, v); // Should be allowed return map2; } Suppose bar is called with U and V as some subtypes of a common supertype that isn't Object: Map<...,...> x = bar(new Sub1(), new Sub2()); U is inferred as Sub1 and V as Sub2. For bar's own call to foo, a good lub would calculate T=lub(U,V)=U (=Sub1), so foo will return Map<U,U> (=Map<Sub1,Sub1>) which is safe to assign to map2 and return to the call ofer bar, which can safely expect a Map<Sub1,Sub1>.
02-11-2006
EVALUATION It is ironic that the result is Object, since the Discussion on page 463 for using *erased* supertypes is precisely to avoid ending up with a lub of Object.
01-11-2006