为什么 std::bitset 以小端方式公开位?

Why does std::bitset expose bits in little-endian fashion?(为什么 std::bitset 以小端方式公开位?)

问题描述

当我使用 std::bitset<N>::bitset( unsigned long long ) 时，它会构造一个位集，当我通过 operator[] 访问它时，这些位似乎以小端方式排序.示例:

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];

打印 1100 而不是 0011 即结束(或 LSB)位于小(低)地址，索引 0.

prints 1100 instead of 0011 i.e. the ending (or LSB) is at the little (lower) address, index 0.

查看标准，它说

将前M位位置初始化为val

程序员自然会想到从 LSB 到 MSB(从右到左)的二进制数字.所以前 M 位位置可以理解为 LSB → MSB，所以位 0 将位于 b[0].

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].

然而，在转变的情况下，定义是这样的

However, under shifting, the definition goes

E1的值<<<E2 是 E1 左移 E2 位位置；空出的位用零填充.

The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-filled.

这里必须将 E1 中的位解释为从 MSB → LSB 然后左移 E2 次.如果它是从 LSB → MSB 写的，那么只有右移 E2 次会得到相同的结果.

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.

令我惊讶的是，在 C++ 的其他任何地方，该语言似乎都投射出自然(英语；从左到右)的书写顺序(在进行移位等按位运算时).为什么这里不一样?

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?

推荐答案

就标准而言，没有字节序的概念.说到std::bitset，[template.bitset]/3定义了bit position:

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:

在 bitset 类的对象和值之间进行转换时一些整数类型，位位置 pos对应位值1<.两位或多位对应的整数值为和它们的位值.