Lambda表达式和泛型方法

Question

interface MyComparable<T extends Comparable<T>>  {
    public int compare(T obj1, T obj2);
}

public static <T extends Comparable<T>> 
       void sort(List<T> list, MyComparable<T> comp) {
    // sort the list
}

List<String> list = Arrays.asList("a", "b", "c");
sort(list, (a, b) -> a.compareTo(b));

interface MyComparable {
    public <T extends Comparable<T>> int compare(T obj1, T obj2);
}

public static <T extends Comparable<T>> 
       void sort(List<T> list, MyComparable comp) {
}

List<String> list = Arrays.asList("a", "b", "c");
sort(list, (a, b) -> a.compareTo(b));

它不编译。它在lambda表达式中显示错误：

“目标方法是泛型的”

好的，当我使用javac编译它时，它显示了以下错误：

SO.java:20: error: incompatible types: cannot infer type-variable(s) T#1
        sort(list, (a, b) -> a.compareTo(b));
            ^
    (argument mismatch; invalid functional descriptor for lambda expression
      method <T#2>(T#2,T#2)int in interface MyComparable is generic)
  where T#1,T#2 are type-variables:
    T#1 extends Comparable<T#1> declared in method <T#1>sort(List<T#1>,MyComparable)
    T#2 extends Comparable<T#2> declared in method <T#2>compare(T#2,T#2)
1 error

sort(list, <T extends Comparable<T>>(a, b) -> a.compareTo(b));

public static <T extends Comparable<T>> int compare(T obj1, T obj2) {
    return obj1.compareTo(obj2);
}

sort(list, SO::compare);

Answer 1

如果函数接口中的方法具有类型参数，则不能对函数接口使用lambda表达式。参见JLS8第15.27.3节：

lambda表达式是兼容的[..]对于目标类型T，如果T是函数接口类型(§9.8)，并且表达式与[..]的函数类型一致T.[..]如果下列所有条件均为真，则lambda表达式与函数类型一致：

• 函数类型没有类型参数。
• [..]