Question

If I have the following method:

public <U extends Number> void doSomething(List<U> l){      
}  

Then due to type erasure the compiler will make it to doSomething(List<Number> l). Right?
If this is the case, then why it is not possible to declare the following along with this:

public void doSomething(List<?> l){  
}

Isn’t this second method, type erased to doSomething(List<Object> l)? Why do I get compiler error of same erasure for these 2 methods?

Answer 1

Your thinking is wrong. Erasure leads to both methods having this signature (List argument types being erased):

public void doSomething(List l) {  
}

Hence, the collision. What you thought was possible to do is this:

public <U extends Number> void doSomething(U argument) {      
}  
public <U extends Object> void doSomething(U argument) {      
}

In this case, after erasure, the method signatures will become this (after U having been erased)

public void doSomething(Number argument) {      
}  
public void doSomething(Object argument) {      
}  

In this case, there is no signature collision.