如何生成的8位唯一号码？ [英] How to Generate Unique Number of 8 digits?

问题描述

我使用此代码生成一个8位数的唯一编号。

I am using this code to generate a 8 digit unique number.

byte[] buffer = Guid.NewGuid().ToByteArray();
return BitConverter.ToUInt32(buffer, 8).ToString();

这段代码真正生成一个唯一的编号，还是会再次重复同样的号码吗？

Does this code really generate a unique number or might it repeat the same number again?

推荐答案

一个GUID不只是一个随机数;它的组成部分的。一些段，如果在相同的计算机上生成的GUID将不会发生任何变化。通过使用你正在打破的GUID的结构和最有可能打破所生成的数字的唯一性的原始的128位的唯一的64位。

A GUID is not just a random number; it's composed of segments. Some of the segments will not change at all if the guid is generated on the same computer. By using only 64-bits of the original 128-bits you are breaking the structure of the guid and most likely breaking the uniqueness of the generated number.

这问题对GUID的独特性更多信息，请查看此链接以及更多的信息，为什么它的坏只使用一个，如果你需要一个唯一的编号GUID。

This question has more info on uniqueness of guids, check this link as well for more info on why it's bad to use only part of a guid if you need a unique number.

如果你需要重复限制降到最低，增量计数器会给你你所需要的。如果您的应用程序使用多个线程或进程，计数器可能很难（甚至不可能）正确地执行。

If you need to limit duplication to an absolute minimum, an incremental counter will give you what you need. If your application is using multiple threads or processes, a counter may be hard (or even impossible) to implement correctly.

这是GUID的被设计为，为的境界可以在多台机器唯一的。因此，如果跨机器的独特性是一个要求，你应该使用的GUID。整个GUID。

This is the realm that guids were designed for, to be unique across multiple machines. So if uniqueness across machines is a requirement you should use guids. The whole guid.

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