勘误：《面经手册》，第3篇《HashMap核心知识》 关于扰动函数图像对比的比较

[JustShowTime]

问题：原文中指出：从这两种的对比图可以看出来，在使用了扰动函数后，数据分配的更加均匀了 答案：其实不然，因为对比的坐标系不同，见下图 总结：至少从上述数据看，扰动函数并不能使数据分配的更加均匀 疑惑：那扰动函数的作用到底是啥呢？单个数据能否认作者数据分配的更加均匀的观点么？以后还需要继续考究

[xiaochen0517]

是这样的的，我在实际运行绘图时也发现了这个问题，统一坐标系之后是无法区分是否扰动的。

[JustShowTime]

谢谢回复，我都快忘了这件事了。。。看到回复我又看了下官方文档，找到了这么一段注释 Because many common sets of hashes are already reasonably distributed (so don't benefit from spreading) 因为许多常见的散列集已经被合理地分配（所以不能从传播中受益）

因此：我觉得最初确实是为了解决hash分布更均匀的问题，但是目前大多数的hash计算已经很均匀了，因此其结果扰动后，影响不大（甚至就会出现和原来相比，略微不均匀的情况）， 举个例子（不一定很恰当）： 磨东西：以前hash之后的，不是很平，加上磨砂纸（扰动函数）会把物件磨得很平；但是呢，现在的hash已经把物件磨得很平，很均匀了，就像镜子一样，然后你又用了磨砂纸（扰动函数），把镜子就糊了。。。 源码如下： /** * Computes key.hashCode() and spreads (XORs) higher bits of hash * to lower. Because the table uses power-of-two masking, sets of * hashes that vary only in bits above the current mask will * always collide. (Among known examples are sets of Float keys * holding consecutive whole numbers in small tables.) So we * apply a transform that spreads the impact of higher bits * downward. There is a tradeoff between speed, utility, and * quality of bit-spreading. Because many common sets of hashes * are already reasonably distributed (so don't benefit from * spreading), and because we use trees to handle large sets of * collisions in bins, we just XOR some shifted bits in the * cheapest possible way to reduce systematic lossage, as well as * to incorporate impact of the highest bits that would otherwise * never be used in index calculations because of table bounds. */ static final int hash(Object key) { int h; return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16); }

[xiaochen0517]

感谢，我明白了。