java中的引用类型大小 [英] reference type size in java

问题描述

为什么java中的引用类型占用8个字节？为什么不少于或多于8个字节？

Why does a reference type in java take 8 bytes? Why not less or more than 8 bytes?

推荐答案

实际上，它没有指定引用变量应该有多少字节，并且事实上，它并不是无处不在。

Actually, it is nowhere specified how much bytes a reference variable shall have, and in fact, it is not everywhere the same.

32位系统的常见虚拟机（即内存总线中具有32位地址的系统）通常使用32位（= 4个字节，与 int 和 float 相同）作为对象引用的大小，而64-的虚拟机位系统通常使用64位（= 8字节）的本机地址。 （请注意，大多数64位系统也可以运行32位程序，所以即使在那里你也常常使用32位VM。）

Common virtual machines for 32-bit systems (i.e. systems with 32 bit adresses in the memory bus) usually use 32-bit (= 4 bytes, same as int and float) as the size for object references, while virtual machines for 64-bit systems often use the native address size 64 bits (= 8 bytes) for these. (Note that most 64 bit systems can run 32-bit programs, too, so often even there you'll be using a 32 bit VM.)

它只是一个简化实现的问题，如果你可以使用实际的内存地址作为你的引用而不是别的东西。

It is simply a matter of simplifying the implementation, if you can use an actual memory address for your references instead of something else.

因为这增加了所用内存的大小（我们经常实际上需要访问 更多的内存），来自Java 7，在64位HotSpot VM上能够在某些条件下使用32位引用，即当堆小于32 GB时（8· 2 ³² 字节）。要将它们转换为实际的内存地址，它们将乘以8（因为对象将在8字节边界上对齐），然后将其添加到基址（如果这不是零）。 （对于小于4 GB的堆，我们不需要乘法步骤。）

As this increases the size of memory used (and often we don't actually need to access that much memory), from Java 7 on the 64-bit HotSpot VM will be able to use 32-bit references under certain conditions, i.e. when the heap is smaller than 32 GB (8·2³² bytes). To convert them to a actual memory address, they are multiplied by 8 (as objects will be aligned on 8-byte-boundaries) and then added to the base address (if this is not zero). (For heaps smaller than 4 GB, we don't need the multiplication step.)

其他虚拟机可能使用类似的技巧。

Other VMs might use similar tricks.

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