Get the method name from a java.util.function.Function

Question

Is it possible to get the method name of a java.util.function.Function. I would like to log each time the name of the method being used. The following example prints the Lambda object but I have not found an easy way to get the method name:

public class Example {

    public static void main(String[]  args) {
        Example ex = new Example();
        ex.callService(Integer::getInteger, "123");
    }

    private Integer callService(Function<String, Integer> sampleMethod, String input) {
            Integer output = sampleMethod.apply(input);
            System.out.println("Calling method "+ sampleMethod);
            return output;
    }
}

I doubt there is one. What name do you expect if someone just passes in a lambda, what name should that expression have? — luk2302
– luk2302, Commented Nov 15, 2017 at 16:09
There is no way, as the lambda, which you're passing is almost identical to an anonymous class implementation which just calls a method in its overriden method — Lino
– Lino, Commented Nov 15, 2017 at 16:14
I would expect that somehow the Lambda expression holds this kind information and print the method name: getInteger() — Georgios Stathis
– Georgios Stathis, Commented Nov 15, 2017 at 16:14
@GeorgiosStathis, well, it doesn't have this information. What if the lambda you got is something along the lines of i -> i.getMyValue().getInternal("Parameter")? What name would you expect to see? — M. Prokhorov
– M. Prokhorov, Commented Nov 15, 2017 at 16:41
Hmm.. I should have studied this better. I expected that since inside the samplemethod object the apply method knows to which method to use the input parameter then the method name should be retrievable somehow. — Georgios Stathis
– Georgios Stathis, Commented Nov 15, 2017 at 17:09

Lino · Accepted Answer · 2017-11-15 16:21:02Z

5

You have to think about, what you're actually doing, when passing a method reference. Because this:

Integer::getInteger

Is almost identical (there is a stack layer more with the below approach) to this:

s -> Integer.getInteger(s)

And above is again similar to the following:

new Function<String, Integer> {
    @Override
    public Integer apply(String s){
        return Integer.getInteger(s);
    }
}

And in the last snippet you clearly see that there is no logical connection to the called method Integer#getInteger(String). Which explaind why it is impossible to do what you intend without introducing something new.

answered Nov 15, 2017 at 16:21

Lino

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