Question

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.

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?

This is under VS2010.

Answer 1

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;

And then you can do this, which is also very fast:

int word = m_pStorage[wordIndex];
bool bit = ((word & (1 << bitIndex)) >> bitIndex) == 1;

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.

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.