如果我想在不同状态下应用不同的操作，如何避免编译错误 "lambda expression should be final or effectively final"？

Question

例如我有两个尽可能简单的类，A和B 我想对 B 的对象采取一些操作，如果 A 对象的某些特定字段发生变化，我应该做一件事，如果其他字段发生变化，我应该做第二件事，我如何使用 Lambda 做到这一点？

答：

 public class A {
        private int someField;
        private String nextField;
    
        public A(int someField, String nextField) {
            this.someField = someField;
            this.nextField = nextField;
        }
    
        public int getSomeField() {
            return someField;
        }
    
        public void setSomeField(int someField) {
            this.someField = someField;
        }
    
        public String getNextField() {
            return nextField;
        }
    
        public void setNextField(String nextField) {
            this.nextField = nextField;
        }
    }

乙：

public class B {
    private String someField;

    public String getSomeField() {
        return someField;
    }

    public void setSomeField(String someField) {
        this.someField = someField;
    }

    public B(String someField) {
        this.someField = someField;
    }
}

演示：

 public class Demo {
        public static <T> boolean isFieldChanged(T oldValue, T newValue) {
            return !Objects.equals(oldValue, newValue);
        }
        public static void someActionOne(B test){
            return;
        }
        public static void someActionTwo(B test){
            return;
        }
        public static void main(String[] args) {
            A oldData = new A(35, "old");
            A clientData = new A(25, "ClientData");
            Consumer<B> action = null;
            if (isFieldChanged(oldData.getNextField(), clientData.getNextField())) {
                action = Demo::someActionOne;
            } else if (isFieldChanged(oldData.getSomeField(), clientData.getSomeField())) {
                action = Demo::someActionTwo;
            }
            List<B> mockData = new ArrayList<>(Arrays.asList(new B("test1"), new B("test2")));
            mockData.forEach(b -> action.accept(b));
        }
    }

在那种情况下如何避免编译错误？

Answer 1

要成为effectively-final，变量在初始化后不得更改。

如果你想使用不同的动作，只需将它们初始化两次：

    public static void main(String[] args) {
        A oldData = new A(35, "old");
        A clientData = new A(25, "ClientData");
        List<B> mockData = new ArrayList<>(Arrays.asList(new B("test1"), new B("test2")));
        if (isFieldChanged(oldData.getNextField(), clientData.getNextField())) {
            mockData.forEach(Demo::someActionOne);
        } else if (isFieldChanged(oldData.getSomeField(), clientData.getSomeField())) {
            mockData.forEach(Demo::someActionTwo);
        }
    }

如果我想在不同状态下应用不同的操作，如何避免编译错误 "lambda expression should be final or effectively final"？

How can I avoid complie error "lambda expression should be final or effectively final", if I want to apply different actions under different state?

java

lambda

if-statement

effectively-final