为什么递归调用此函数不会抛出NullPointerException [英] Why calling this function recursively does not throw a NullPointerException

问题描述

我的问题来自主题。

考虑以下代码：

public class Test {    
    static Function<Integer, Integer> fibLambda = null;
    public static void main (String[] args) { 
        fibLambda = n -> n <= 2 ? 1 : fibLambda.apply(n - 1) + fibLambda.apply(n - 2); 
        System.out.println(fibLambda.apply(6));
    }
}

上面的输出是8.

我没有得到的是 fibLamdba 是如何初始化的？似乎我完全错过了方法调用是如何完成的，因为我虽然这段代码会产生一个NPE。

What I don't get is that how fibLamdba is initialized? It seems that I totally miss how the method invocation is done because I though that this code would produce a NPE.

希望我的问题很清楚

推荐答案

您的代码相当于

static Function<Integer, Integer> fibLambda = null;

public static void main(String[] args) {
    fibLambda = n -> n <= 2 ? 1 : Example.fibLambda.apply(n - 1) + Example.fibLambda.apply(n - 2);
    System.out.println(fibLambda.apply(6));
}

当适用时被称为 fibLambda 被赋值。基本上，lambda表达式不捕获 fibLambda 的值，它只是注册该变量需要在适当的时刻进行评估以产生值。

By the time the apply is called fibLambda is assigned a value. Basically, the lambda expression doesn't capture the value of fibLambda, it just registers that the variable needs to be evaluated at the appropriate moment to produce a value.

请记住，lambda表达式不会执行其正文中出现的代码。它只是一个声明，类似于声明匿名类实例的方式。

Remember that a lambda expression doesn't execute the code appearing in its body. It's just a declaration, similar to how you declare an anonymous class instance.

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