感觉 Lambda 加上递归调用,就有点搞不懂运行过程了 - V2EX
amiwrong123
V2EX    Java

感觉 Lambda 加上递归调用,就有点搞不懂运行过程了

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  •   amiwrong123 Jul 19, 2020 2622 views
    This topic created in 2123 days ago, the information mentioned may be changed or developed.

    https://bugs.openjdk.java.net/browse/JDK-8240162

    这个 ConcurrentHashMap 的 bug,运行它的用例,会产生一个 bug,就是明明 map 中有映射,但 size 为 0 。

    import java.util.concurrent.*; public class test5 { public static void main(String[] args) { final ConcurrentHashMap<String, String> map = new ConcurrentHashMap<>(); final String key = "key"; final String value = "value"; map.put("anotherKey", "anotherValue"); map.put(key, value); System.out.println("size=" + map.size()); map.computeIfPresent(key, (aKey, aValue) -> { map.computeIfPresent(key, (bKey, bValue) -> null); return null; }); System.out.println("size=" + map.size());//这句返回 0 map.put(key, value);//这句打断点 System.out.println("size=" + map.size()); } } 
    bug

    可以发现明明 map 中有映射,但 size 为 0 。

    1. 为什么没有产生无限递归,看起来会无限递归啊?感觉调用过程好难理解
    2. 为啥执行完,size 返回错误的值?

    运行环境 jdk8

    5 replies    2020-07-20 19:33:00 +08:00
    amiwrong123
        1
    amiwrong123  
    OP
       Jul 19, 2020 via Android
    好吧,看错了,不会产生无限递归,不过咋一看这个过程确实有点搞不清楚
    312ybj
        2
    312ybj  
       Jul 19, 2020 via Android
    你可以尝试改变 debug 的类型,右击一下短点
    yanshenxian
        3
    yanshenxian  
       Jul 19, 2020   1
    ConcurrentHashMap 的 size 方法本来就是不精确的
    这个例子也没有递归操作啊 computeIfPresent 里的 lambda 返回空相当于 remove 这个 key 了,内部操作 count -1
    这就相当于操作了两次 -1 最后计算的 size 就是 0

    如果换成 HashMap 这里就会抛出 ConcurrentModificationException
    no1xsyzy
        4
    no1xsyzy  
       Jul 20, 2020
    没有哪个函数调用了自身,只是存在某处回调中再次调用同样的外部函数,你可能把它误看作是的递归了。
    amiwrong123
        5
    amiwrong123  
    OP
       Jul 20, 2020 via Android
    @no1xsyzy
    是的,并不是递归
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