将字节拆分为8个数字 [英] Split byte into 8 numbers

问题描述

您好，是否可以以0到31的单字节存储8个整数？

我理解如何使用位移将一个32位整数分割成4个字节，

但是我不知道单字节是否可以。

一个字节可容纳1个数字，最多255个。如果是2个数字，

我假设它们可能达到127 ...如果是8，看起来喜欢高达31

应该是真实的。



或者我完全错了，这是不可能的？



我想有类似的东西：

Hello, is it possible to store 8 integers in range 0 - 31 in single byte?
I understand how to split one 32 bit integer into 4 bytes using bit shifting,
but I don't know if the same is possible with single byte.
One byte can hold 1 number up to 255. In case of 2 numbers,
I a assume they could be up to 127... in case of 8, looks like up to 31
should be real.

Or am I completely wrong and this is not possible?

I would like to have something like:

int getAt(int pos, byte b) // pos is 0 - 7
{
   // extraction code
}

void setAt(int pos, byte & b) // pos is 0 - 7
{
   // insertion code
}

我有什么尝试过：



我只在网上搜索并且像b>>一样移动1，但没有运气。

What I have tried:

I have tied only to search on the web and shifting like b >> 1, but without luck.

推荐答案

当然你可以在一个字节中存储8个整数。但那些整数只能有0或1的值。

of course you can store 8 integers in a byte. but those integers can only have values 0 or 1.

Quote：

或者我完全错了，这是不可能的？

Or am I completely wrong and this is not possible?

这是不可能的。

要编码0-31之间的数字，你需要5位，所以编码8它们以一种你可以检索它们的方式，你需要40位。

It is not possible.
To encode a number between 0-31, you need exactly 5 bits, so to encode 8 of them in a way you can retrieve them, you need 40 bits.

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