Is java actually overloading methods based on generic when lambda involved? -

i totally understand erasure of generic , therefore java cant overloading base on generic. if take lambda expression consideration?

see following case.

interface base{     void test(); } // compiler cant distinguish these 2 interfaces lambda expression provided  interface foo extends base{} interface boo extends base{}  public static void main(string[] args) throws classnotfoundexception {     hashmap<string,long> set = new hashmap<>();     // valid, since type foo explicitly given out     t1(set,new foo(){         @override         public void test(){         }     }); // "long"      // using lambda expression make second argument indistinguishable,      //so overloading can depend on first argument     //hashmap generic type.     // compiler still managed work out     t1(set,()->{});// "long"      hashmap<string,integer> set2 = new hashmap<>();     t1(set2,()->{});// "integer" } public static void t1(hashmap<string,long> set,foo foo){system.out.println("long");} public static void t1(hashmap<string,integer> set,boo boo){system.out.println("integer");} 

the result of each function call expected without compile or runtime error. , that's weired part of whole thing: function overloading based on generic types.

so going on behind scenes?

interesting question.

the compiler decides @ compile time type lambda function is, though generics wiped away @ point, type of function set @ point, meaning knows method has called.

my guess can because still has generic information @ stage -- i.e. compiler gets rid of it.

if @ bytecode class says:

16: invokestatic  #25                 // method t1:      (ljava/util/hashmap;ltest$foo;)v 25: invokestatic  #25  // method t1:        (ljava/util/hashmap;ltest$foo;)v 

so hashmap typing gone, interface, type has been set.