如何转换为的BitSet字节

如何转换为的BitSet字节

本文介绍了如何转换为的BitSet字节/ UINT8数组?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我需要从可能(不)包含CHAR_BIT位的倍数位集extact字节。我现在有多少位集的位,我需要投入一个数组。例如,

I need to extact bytes from the bitset which may (not) contain a multiple of CHAR_BIT bits. I now how many of the bits in the bitset I need to put into an array. For example,

设置声明位为和std :: bitset< 40 GT; ID;

有一个单独的变量 NBITS 有多少位的 ID 是可用的。现在,我想在CHAR_BIT的倍数来提取这些位。我也需要照顾的情况下,其中 NBITS%CHAR_BIT!= 0 。我好把这个变成UINT8数组

There is a separate variable nBits how many of the bits in id are usable. Now I want to extract those bits in multiples of CHAR_BIT. I also need to take care of cases where nBits % CHAR_BIT != 0. I am okay to put this into an array of uint8

推荐答案

不幸的是还有的语言中没有什么好办法,假设你需要比位在数无符号长(在这种情况下,你可以使用 to_ulong )。你必须遍历所有的位,并生成字节数组自己。

Unfortunately there's no good way within the language, assuming you need for than the number of bits in an unsigned long (in which case you could use to_ulong). You'll have to iterate over all the bits and generate the array of bytes yourself.

这篇关于如何转换为的BitSet字节/ UINT8数组?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!

07-29 18:00