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.

Answer 2

There is no notion of endian-ness as far as the standard is concerned. When it comes to std::bitset, [template.bitset]/3 defines bit position:

When converting between an object of class bitset<N> and a value of some integral type, bit position pos corresponds to the bit value 1<<pos. The integral value corresponding to two or more bits is the sum of their bit values.

Using this definition of bit position in your standard quote

initializing the first M bit positions to the corresponding bit values in val

a val with binary representation 11 leads to a bitset<N> b with b[0] = 1, b[1] = 1 and remaining bits set to 0.