When I use std::bitset<N>::bitset( unsigned long long ) this constructs a bitset and when I access it via the operator[], the bits seems to be ordered in the little-endian fashion. Example:
std::bitset<4> b(3ULL);
std::cout << b[0] << b[1] << b[2] << b[3];
prints 1100 instead of 0011 i.e. the ending (or LSB) is at the little (lower) address, index 0.
Looking up the standard, it says
initializing the first M bit positions to the corresponding bit values in
val
Programmers naturally think of binary digits from LSB to MSB (right to left). So the first M bit positions is understandably LSB → MSB, so bit 0 would be at b[0].
However, under shifting, the definition goes
The value of
E1<<E2isE1left-shiftedE2bit positions; vacated bits are zero-filled.
Here one has to interpret the bits in E1 as going from MSB → LSB and then left-shift E2 times. Had it been written from LSB → MSB, then only right-shifting E2 times would give the same result.
I'm surprised that everywhere else in C++, the language seems to project the natural (English; left-to-right) writing order (when doing bitwise operations like shifting, etc.). Why be different here?
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 positionposcorresponds to the bit value1<<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
Mbit positions to the corresponding bit values inval
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.