Why does std::bitset expose bits in little-endian fashion?

Question

When I use std::bitset::bitset( unsigned long long ) this constructs a bitset and when I access it via the operator[], the bits seems to b

Answer 1

This is consistent with the way bits are usually numbered - bit 0 represents 2⁰, bit 1 represents 2¹, etc. It has nothing to do with the endianness of the architecture, which concerns byte ordering not bit ordering.