方法参考静态与非静态

Question

I've wondered how to distinct between static and non static method references with the same name. In my example I have a class called StringCollector that has the following three methods:
StringCollector append(String string)
static StringCollector append(StringCollector stringCollector, String string)
StringCollector concat(StringCollector stringCollector)
Now if I want to use a Stream<String> to collect a list of strings I would write something like that:
Arrays.asList("a", "b", "c").stream()
.collect(StringCollector::new, StringCollector::append, StringCollector::concat);
As we can see the code doesn't compile. I think that's because the compiler can't deside, which method to use because each of them would match the functional. The question is now: Is there any possible way to distinct static method references from instance method references?

(PS: Yes the code compiles if I rename one of the two methods. For each of them.)

Answer 1

In this case unbound reference to the instance method append has the same arity, argument types and even return value as the reference to the static method append , so no, you cannot resolve the disambiguation for method references. If you don't want to rename one of the methods, you should use lambda instead:

collect(StringCollector::new, (sb, s) -> sb.append(s), StringCollector::concat);

Or if you actually want to use static method:

collect(StringCollector::new, (sb, s) -> StringCollector.append(sb, s),
        StringCollector::concat);