Why does a lambda change overloads when it throws a runtime exception?

Question

Bear with me, the introduction is a bit long-winded but this is an interesting puzzle.

I have this code:

public class Testcase {
    public static void m         


        

Answer 1

First, according to §15.27.2 the expression:

() -> { throw ... }

Is both void-compatible, and value-compatible, so it's compatible (§15.27.3) with Supplier>:

class Test {
  void foo(Supplier> bar) {
    throw new RuntimeException();
  }
  void qux() {
    foo(() -> { throw new IllegalArgumentException(); });
  }
}

(see that it compiles)

Second, according to §15.12.2.5 Supplier (where T is a reference type) is more specific than Runnable:

Let:

• S := Supplier
• T := Runnable
• e := () -> { throw ... }

So that:

• MTs := T get() ==> Rs := T
• MTt := void run() ==> Rt := void

And:

• S is not a superinterface or a subinterface of T
• MTs and MTt have the same type parameters (none)
• No formal parameters so bullet 3 is also true
• e is an explicitly typed lambda expression and Rt is void