JavaLambdasAndMethodReferences.java

    private int methodWithoutParameter() {
        return 0;
    }

    /**
     * Equals exactly method signature of java.util.function.Function:apply
     */
    private int methodWithParameter(int x) {
        return x;
    }

    @Test
    public void blubber() {

        Map<Integer, Integer> myMap = new HashMap<>();

        // Both lambda and method reference can be used to implement Function, if the method signature equals
        // signature of java.util.function.Function::apply:
        Function<Integer, Integer> functionWithParameter1 = (k) -> methodWithParameter(k);
        Function<Integer, Integer> functionWithParameter2 = this::methodWithParameter;
        myMap.computeIfAbsent(2, functionWithParameter1);
        myMap.computeIfAbsent(2, functionWithParameter2);
        functionWithParameter1.apply(5);

        // Only lambda can be used to implement Function if the method signature differs from the right signature:
        Function<Integer, Integer> functionWithoutParameter1 = k -> methodWithoutParameter();
        Function<Integer, Integer> functionWithoutParameter2 = this::methodWithoutParameter; // not possible
        myMap.computeIfAbsent(2, functionWithoutParameter1);
        functionWithoutParameter1.apply(5);
    }

@schauder Das hier meine ich

The types don't match. A method without parameter returning an int is a Supplier<Integer>. A lambda taking an int argument, ignoring it and returning an int is still a Function<Integer, Integer.

If you want a lambda equivalent to a method reference of a method without parameter you would want

() -> methodWithoutParameter()

And you wouldn't be able to assign that to a Function<Integer, Integer> either.

Great! Wrote an article about that.