[英]Why calling this function recursively does not throw a NullPointerException
My question comes from this thread . 我的问题来自这个话题 。
Consider this code: 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));
}
}
The output above is 8. 上面的输出是8。
What I don't get is that how fibLamdba
is initialized? 我不知道如何初始化
fibLamdba
? It seems that I totally miss how the method invocation is done because I though that this code would produce a NPE. 似乎我完全想念方法调用是如何完成的,因为我虽然这段代码会产生NPE。
Hope my question is clear 希望我的问题清楚
Your code is equivalent to 您的代码等同于
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));
}
By the time the apply
is called fibLambda
is assigned a value. 到名为
fibLambda
的apply
fibLambda
被分配一个值时。 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. 基本上,lambda表达式不会捕获
fibLambda
的值,它只是注册该变量需要在适当的时候进行求值才能产生一个值。
Remember that a lambda expression doesn't execute the code appearing in its body. 请记住,lambda表达式不会执行其主体中出现的代码。 It's just a declaration, similar to how you declare an anonymous class instance.
这只是一个声明,类似于声明匿名类实例的方式。
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