Is
test::testMethod == test::testMethod
true? I want to find out whether they will refer to the same object. This code does not compile. But there may be several situations in which this needs clarification.
I suspect that this will expand to
Runnable r = () -> test.testMethod()
Runnable r1 = () -> test.testMethod()
And whether the below is true.
r == r1
even if your code could compile and the test::testMethod declared as Runnable the answer test::testMethod == test::testMethod is always return false, because you comparing with two diff class instances. for each lambda expression the compiler will create a synthentic anonymous inner class for it.for example:
//a synthentic anonymous lambda class A implemented Runnable
Runnable r = () -> test.testMethod();
//a synthentic anonymous lambda class B implemented Runnable
Runnable r1 = () -> test.testMethod();
r.getClass().equals(r1.getClass());// always return false
Let's look at the following example:
Predicate<String> predicate = String::isEmpty;
Function<String,Boolean> function = String::isEmpty;
System.out.println(predicate.equals(function)); // false
A lambda doesn't contain any information about it's type. That information is deduced from the context. The same lamba String::inEmpty can represent different functional interfaces, as we have seen above.
trueor not is meaningless. Please provide a complete example that does compile.true, but doesn’t have to. And in the current JRE, it isn’t. You may also retrace the test of this answer which shows, under which circumstances the objects are the same in the current implementation.