Java Lambda not compiling

Question

Can someone tell why doReduce compiles but doReduce2 doesn't:

List<String> strings = List.of("a", "bb", "cc", "ddd");
Map<Integer,ObjStr2> result2 = strings.stream()
                .map(ObjStr1::new)
                .collect(Collectors.groupingBy(ObjStr1::getLen
                        , Collectors.reducing(new ObjStr2(), ObjStr1::to2, ObjStr2::doReduce)));

class ObjStr1{
    String str = "";

    ObjStr1(String str) {
        this.str = str;
    }

    static ObjStr2 to2(ObjStr1 o){
        return new ObjStr2(o.str);
    }

    Integer getLen(){return str.length(); };
}

class ObjStr2{
    String str = "";

    ObjStr2(){}

    ObjStr2(String str) {
        this.str = str;
    }

    static BinaryOperator<ObjStr2> doReduce2(){
        return (a,b) -> new ObjStr2(a.str + b.str);
    }

    static ObjStr2 doReduce(ObjStr2 a, ObjStr2 b){
        return new ObjStr2(a.str + b.str);
    }

    @Override
    public String toString(){
        return str;
    }
}

I can even copy the contents of doReduce2 directly into the lambda and it will compile.
reducing is supposed to accept BinaryOperator

@Andreas he means this: return (a,b) -> new ObjStr2(a.str + b.str); — Stultuske
– Stultuske, Commented Apr 24, 2020 at 8:38
Surely you're reducing a stream of ObjStr1? So you can't apply a BinaryOperator<ObjStr2> to it? — Andy Turner
– Andy Turner, Commented Apr 24, 2020 at 8:39

Naman · Accepted Answer · 2020-04-24 08:50:40Z

1

You should try calling the method doReduce2 without the use of method reference and it should work -

Map<Integer, ObjStr2> result2 = strings.stream()
        .map(ObjStr1::new)
        .collect(Collectors.groupingBy(ObjStr1::getLen,
                Collectors.reducing(new ObjStr2(),
                        ObjStr1::to2, ObjStr2.doReduce2())));

edited Apr 24, 2020 at 8:50

answered Apr 24, 2020 at 8:44

Naman

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Andreas · Accepted Answer · 2020-04-24 08:49:18Z

1

doReduce() is a BinaryOperator.

doReduce2() returns a BinaryOperator.

Since reduce() expects a BinaryOperator, you can give doReduce as a method reference, or you can call doReduce2().

answered Apr 24, 2020 at 8:49

Andreas

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