如何转换为的BitSet字节/ UINT8数组？ [英] How do I convert bitset to array of bytes/uint8?

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＆LT; 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.

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