Java var与推理类型歧义

Question

Collectors.groupingBy((String s)->s.toLowerCase(),Collectors.counting());
Collectors.groupingBy((String s)->s.toLowerCase(Locale.ENGLISH),Collectors.counting());

Collectors.groupingBy(String::toLowerCase,Collectors.counting());

当然，如果我给IntelliJ添加更多上下文，比如：

Collector<String,?,Map<String,Long>> c = Collectors.groupingBy(String::toLowerCase,Collectors.counting());

这是正确的，但在Java10 var推断类型上下文中，这是错误的：

var c = Collectors.groupingBy(String::toLowerCase,Collectors.counting());

我理解编译器不能推断counting的输入类型。如果我写：

Collector<String,?,Long> counter = Collectors.counting();
var c = Collectors.groupingBy(String::toLowerCase,counter);

Reference to 'toLowerCase' is ambiguous, both 'toLowerCase(Locale)' and 'toLowerCase()' match

编译器的错误更加不可读（但包含更多关于推理失败原因的提示），如下所示：

Demo.java:31: error: incompatible types: cannot infer type-variable(s) T#1,K,A,D,CAP#1,T#2
        Collectors.groupingBy(String::toLowerCase,Collectors.counting());
                             ^
    (argument mismatch; invalid method reference
      incompatible types: Object cannot be converted to Locale)
  where T#1,K,A,D,T#2 are type-variables:
    T#1 extends Object declared in method <T#1,K,A,D>groupingBy(Function<? super T#1,? extends K>,Collector<? super T#1,A,D>)
    K extends Object declared in method <T#1,K,A,D>groupingBy(Function<? super T#1,? extends K>,Collector<? super T#1,A,D>)
    A extends Object declared in method <T#1,K,A,D>groupingBy(Function<? super T#1,? extends K>,Collector<? super T#1,A,D>)
    D extends Object declared in method <T#1,K,A,D>groupingBy(Function<? super T#1,? extends K>,Collector<? super T#1,A,D>)
    T#2 extends Object declared in method <T#2>counting()
  where CAP#1 is a fresh type-variable:
    CAP#1 extends Object from capture of ?

Answer 1

这是编译器的“弱点”，至少在这个JEP到位之前是如此。

我在这里已经回答了几乎相同的问题。JDK核心开发人员也给出了另一个答案。

还有一个问题和你很接近。