Java 32位系统BitSet的内存大小 [英] Memory size of Java 32-bit system BitSets

问题描述

如何在C ++中计算new BitSet(n) 的内存.

How to compute the memory of new BitSet(n) in C++.

内存占用了new BitSet(1024) 在Java中.

What memory takes the new BitSet(1024) in Java.

但是Java的公式似乎有所不同.我想计算用于new BitSet(100000)的内存，请您帮忙?

But it seems the formula for Java is different. I want to compute the memory spent for new BitSet(100000), could you please help?

推荐答案

BitSet被打包到单词"数组中.一个单词(在当前实现中)很长.内部的单个位将被检索/使用掩码进行设置；因此，它在内部将64位压缩为一个long值，并使用long数组来容纳所需的所有位.

BitSet are packed into arrays of "words." A word is (in the current implementation) a long. The single bits inside will be retrieved / set uses masking; therefore it internally packs 64 bits in one single long value, and uses an array of longs to hold all the bits you need.

数组的大小将为N(100000)/64字节，即1563长，即12504字节，再加上BitSet用于其内部结构/布告的固定开销.

The dimension of the array will be N (100000) / 64 bytes, or 1563 longs, or 12504 bytes, plus a fixed overhead needed by BitSet for its internal structure/bookeeping.

请参见 http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/util/BitSet.java ；计算字段并总结所需的空间(一个int:4个字节；一个long:8个字节，依此类推)，您可以了解固定开销是多少.

See http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/util/BitSet.java for the implementation; counting the fields and summing up the space they need (an int: 4 bytes; a long: 8 bytes, and so on) you can understand how much is the fixed overhead.

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