将int stream转换为map

Question

我有一个int流,并希望该流的每个元素进行一些计算,并将它们作为Map返回,其中键是int值,值是该计算的结果.我写了下面这段代码:

IntStream.range(0,10).collect(Collectors.toMap(Function.identity(), i -> computeSmth(i)));

哪里computeSmth(Integer a).我有下一个编译器错误

 method collect in interface java.util.stream.IntStream cannot be applied to given types;
  required: java.util.function.Supplier<R>,java.util.function.ObjIntConsumer<R>,java.util.function.BiConsumer<R,R>
  found: java.util.stream.Collector<java.lang.Object,capture#1 of ?,java.util.Map<java.lang.Object,java.lang.String>>
  reason: cannot infer type-variable(s) R
    (actual and formal argument lists differ in length)

我做错了什么？

Answer 1

这是我的代码,它适合你.

功能参考版本

public class AppLauncher {

public static void main(String a[]){
    Map<Integer,Integer> map = IntStream.range(1,10).boxed().collect(Collectors.toMap(Function.identity(),AppLauncher::computeSmth));
    System.out.println(map);
}
  public static Integer computeSmth(Integer i){
    return i*i;
  }
}

Lambda表达版本

public class AppLauncher {

    public static void main(String a[]){
        Map<Integer,Integer> map = IntStream.range(1,10).boxed().collect(Collectors.toMap(Function.identity(),i->i*i));
        System.out.println(map);
    }
}