Why does Function.identity() break type reification but t -> t does not?

Question

Answers found at Java 8 lambdas, Function.identity() or t->t seem to imply that Function.identity() is almost always equivalent to t -> t. Howeve

Answer 1

Ecj is able to infere the correct(?) type argument (Integer) to match the constraints. Javac for some reason comes to a different result.

Thats not the first time javac/ecj behave differently in inference of type parameters.

In that case you can give javac a hint with Function.<Integer>identity() to make it compileable with javac.

For the difference between Function.identity() and t->t:

• Function.identity() is Function<T,T>
• t->t in that case is Function<? super Integer, ? extends Integer>

So t->t is more flexible in the methods it can match to.