为什么std :: bitset以little-endian方式公开位？

Question

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 ie 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 << E2 is E1 left-shifted E2 bit 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?

Answer 1

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 .

Answer 2

这与位通常被编号的方式是一致的-位0代表2 ⁰ ，位1代表2 ¹ ，依此类推。这与体系结构的字节序无关，后者涉及字节顺序而不是位顺序。