根据整数值使用不同的方法

Question

这是我无法理解的东西,现在我有这样的东西:

boolean method1(int a){
//something
returns true;
}

boolean method2(int a){
//something
returns true;
}

for (int i; i<100; i++){
  switch (someInt){
  case 1: boolean x = method1(i);
  case 2: boolean x = method2(i);
  }


}

我想要的是把开关带出循环,因为someInt将保持相同的每个i,因此需要只决定一次,但我需要x检查每一个,所以我需要像:

    switch (someInt){
          case 1: method1(); //will be used in loop below
          case 2: method2(); //will be used in loop below
          }

   for (int i; i<100; i++){
       boolean x = method the switch above picked
   }

Answer 1

您可以使用Java 8方法引用.

• https://blog.idrsolutions.com/2015/02/java-8-method-references-explained-5-minutes/
• https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html

这是一个例子:

public class WithMethodRefs {
    interface MyReference {
        boolean method(int a);
    }

    boolean method1(int a) {
        return true;
    }

    boolean method2(int a) {
        return false;
    }

    public void doIt(int someInt) {
        MyReference p = null;
        switch (someInt) {
        case 1:
            p = this::method1;
            break;
        case 2:
            p = this::method2;
            break;
        }

        for (int i = 0; i < 100; i++) {
            p.method(i);
        }
    }
}