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线程池线程？(pthread threadpool?)

 pthread库中是否包含一个线程池实现？ 还是有常用的库，人们使用/ 

does the pthread lib include a threadpool implementation? or are there commonly used libs that folks use/

原文：https://stackoverflow.com/questions/4568517
更新时间：2024-01-19 06:01

最满意答案

 由于混合类型，这有点令人费解，但这可以完全用lambdas来完成： 
 

Stream<Object> stream = Stream.concat(list1.stream(), list2.stream());
System.out.println(stream.sorted((Object o1, Object o2) ->
    (o1 instanceof String ? 2*((String)o1).length() : 1+2*(Integer)o1)
    - (o2 instanceof String ? 2*((String)o2).length() : 1+2*(Integer)o2)
  ).collect(Collectors.toList()));

 

 基本的想法是使用标准的compareTo逻辑a - b来定义一个比较长度（如果它是一个字符串）和该值（如果它是一个整数）的lambda。 但是，为了处理一个关系的情况，我将所有值乘以2 ，然后在Integer数值中加上一个额外的1 。 这会导致Integer在排序中最后排序。 
 

 演示： http ： //ideone.com/39ppGD 

This is slightly convoluted because of the mixed types, but here is one way it could be done entirely with lambdas:
 

Stream<Object> stream = Stream.concat(list1.stream(), list2.stream());
System.out.println(stream.sorted((Object o1, Object o2) ->
    (o1 instanceof String ? 2*((String)o1).length() : 1+2*(Integer)o1)
    - (o2 instanceof String ? 2*((String)o2).length() : 1+2*(Integer)o2)
  ).collect(Collectors.toList()));

 

The basic idea is to define a lambda that compares the length (if it's a string) to the value (if it's an integer) using the standard compareTo logic a - b. However, to handle the case of a tie I've multiplied all the values by 2, and then added an extra 1 to the Integer values. This causes Integer to be sorted last in the case of a tie.
 

Demo: http://ideone.com/39ppGD

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