为什么std :: bitset< 8>4个字节大吗? [英] Why is std::bitset<8> 4 bytes big?
问题描述
似乎对于std :: bitset< 1到32>，大小设置为4个字节.对于33到64的大小，它直接跳到8个字节.不会有任何开销，因为std :: bitset< 32>是偶数4个字节.
It seems for std::bitset<1 to 32>, the size is set to 4 bytes. For sizes 33 to 64, it jumps straight up to 8 bytes. There can't be any overhead because std::bitset<32> is an even 4 bytes.
在处理位时，我可以看到与字节长度对齐，但是为什么位集需要与字长对齐，特别是对于最可能在内存预算紧张的情况下使用的容器呢?
I can see aligning to byte length when dealing with bits, but why would a bitset need to align to word length, especially for a container most likely to be used in situations with a tight memory budget?
这是在VS2010下进行的.
This is under VS2010.
推荐答案
我假定通过获取32位值然后隔离相关位来完成对该位集的索引，因为这在处理器指令方面是最快的(有效较小的值在x86上速度较慢).为此所需的两个索引也可以很快地计算出来:
I assume that indexing into the bitset is done by grabbing a 32-bit value and then isolating the relevant bit because this is fastest in terms of processor instructions (working with smaller-sized values is slower on x86). The two indexes needed for this can also be calculated very quickly:
int wordIndex = (index & 0xfffffff8) >> 3;
int bitIndex = index & 0x7;
然后您可以执行此操作，这也非常快:
And then you can do this, which is also very fast:
int word = m_pStorage[wordIndex];
bool bit = ((word & (1 << bitIndex)) >> bitIndex) == 1;
此外，每个位集最多浪费3个字节也不完全是内存问题恕我直言.考虑到位集已经是存储此类信息的最有效的数据结构，因此您必须将浪费作为总结构大小的百分比进行评估.
Also, a maximum waste of 3 bytes per bitset is not exactly a memory concern IMHO. Consider that a bitset is already the most efficient data structure to store this type of information, so you would have to evaluate the waste as a percentage of the total structure size.
对于1025位，此方法将占用132个字节而不是129个字节，从而占用2.3％的开销(并且随着位集站点的增加而减少).考虑到可能的性能优势，这听起来很合理.
For 1025 bits this approach uses up 132 bytes instead of 129, for 2.3% overhead (and this goes down as the bitset site goes up). Sounds reasonable considering the likely performance benefits.
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