为什么Java无法在重载方法中推断类型？

我有以下代码，

javac

无法选择要使用的重载

get（）

方法的版本。为什么不呢

public class x {
    static <T> T get(Key<T> key, T defaultValue)
    {throw new UnsupportedOperationException();}

    static <T> T get(Key<T> key,
            java.util.function.Supplier<T> defaultValueSupplier)
    {throw new UnsupportedOperationException();}

    static void test(){
        Key<String> k = new Key<>();
        String s = get(k, () -> "test");
    }

    private static class Key<T>{}
}

---

然后它就起作用了。是什么导致

javac

将代码与两种方法匹配？

可重复<代码>字符串s=get（k，（供应商）（->“测试”）有效。还有另一个解决方法：

字符串s=x.get（k，（）->“测试”）

@Eran这很有趣，我刚刚用Oracle的1.8.0_60和1.8.0_77（我没有1.8.0_66）进行了测试，也没有编译它。也没有

1.8.0_141

和OpenJDK

11.0.1

-相同的错误消息

% javac -version
javac 1.8.0_202
% javac x.java
x.java:11: error: reference to get is ambiguous
        String s = get(k, () -> "test");
                   ^
  both method <T#1>get(Key<T#1>,T#1) in x and method <T#2>get(Key<T#2>,Supplier<T#2>) in x match
  where T#1,T#2 are type-variables:
    T#1 extends Object declared in method <T#1>get(Key<T#1>,T#1)
    T#2 extends Object declared in method <T#2>get(Key<T#2>,Supplier<T#2>)
x.java:11: error: incompatible types: cannot infer type-variable(s) T
        String s = get(k, () -> "test");
                      ^
    (argument mismatch; String is not a functional interface)
  where T is a type-variable:
    T extends Object declared in method <T>get(Key<T>,T)
2 errors

both method <T#1>get(Key<T#1>,T#1) in x and method <T#2>get(Key<T#2>,Supplier<T#2>) in x match

Key<String> k = new Key<>();
Supplier<String> l = () -> "test";
String s = get(k, l);