为什么不允许`std :: uniform_int_distribution< uint8_t>`和`std :: uniform_int_distribution< int8_t>`？

问题描述

：

如果我不在乎范围，我可以只屏蔽一个更大的类型的位以产生随机数。如果不是，它更复杂。为什么默认情况下不提供字节类型？

If I don't care about the range, I can just mask off the bits of a larger type to generate random numbers. If not, it's more complex. Why aren't the byte types provided just by default?

推荐答案

此，并且其中包括：

建议的解决方案是更改限制以允许标准整数类型：

The proposed resolution is to change the restriction to allow standard integer types:

和：

虽然不是 uint8_t ，但可以获得

This gets you unsigned / signed char although not uint8_t or int8_t but they are likely equivalent. Extended integral types were exluded to simplify the wording and maximize consensus:

注意，这不包括 char 因为它是实现定义 char 是否已签名。

Note, this excludes char since it is implementation defined whether char is signed or not.

标准讨论列表 a>。

Jonathan Wakely指出这个建议有争议，并评论他上次讨论的笔记包括：

Jonathan Wakely notes this proposal is controversial and commented that his notes from the last discussion include the following:

他建议为 random_device 添加一个成员以提供单个字节像一个合理的选择。

He suggests adding a member to random_device to provide single bytes, which is seems like a reasonable alternative.

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