这是一个可为空＆所述的大小;的Int32＆GT ;? [英] What is the size of a Nullable<Int32>?

问题描述

于是，一对夫妇的问题，实际上是：

1. 的 INT （的Int32 ）被指定为（显然）32位。怎么样一个 INT （可空＆LT; INT＆GT; ）？我的直觉告诉我，这将是32位整数加上8个位布尔，但也许是实现比这更复杂。
2. 我会回答用我自己的问题的sizeof（INT？）;但作为诠释？是托管型，这是不允许的。据我所知，一个类型的大小可能是依赖于平台的，而且其中包含对其他对象的对象的情况下，的sizeof 式的操作会产生误导。但是，有没有办法让一个基准的大小（即，什么是新实例化实例的大小会）的托管类型，考虑到目前的环境？

解决方案

您可以在反汇编一看或反思。

如果有两个字段：布尔和 T ，所以大概8个字节（假设4字节对齐）

So, a couple of questions, actually:

1. An int (Int32) is specified to be (obviously) 32 bits. What about an int? (Nullable<int>)? My gut tells me that it would be 32 bits for the integer plus 8 more bits for the boolean, but perhaps the implementation is more intricate than that.
2. I would have answered my own question using sizeof(int?); but as int? is a managed type this is not allowed. I understand that the size of a type may be platform-dependent, and that in the case of objects which contain references to other objects, a sizeof-like operation would be misleading. However, is there a way to get a "baseline" size (i.e., what the size of a newly instantiated instance would be) for a managed type, given the current environment?

解决方案

You can take a look in ildasm or Reflector.

If has two fields: a bool and a T, so probably 8 bytes (assuming 4 byte alignment).

这篇关于这是一个可为空＆所述的大小;的Int32＆GT ;?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！