在编译时初始化c ++ std :: bitset [英] Initializing c++ std::bitset at compile time

问题描述

我正在尝试在编译时使用其某些索引来初始化std :: bitset <256>，例如将50-75和200-225设置为1。

I'm trying to initialize a std::bitset<256> at compile time with some of its indices, lets say 50-75 and 200-225 set to 1.

基于 http://en.cppreference.com/w/ cpp / utility / bitset / bitset
看来我的2个选项是：

Based on http://en.cppreference.com/w/cpp/utility/bitset/bitset It looks like my 2 options are:

constexpr bitset();
constexpr bitset( unsigned long long val ); 

有人考虑第二个构造函数不能很好地解释我如何初始化我的位集吗？

Could someone shed light on how I would initialize my bitset considering the second constructor wouldn't work for large indices?

推荐答案

位集构造函数 constexpr bitset（unsigned long long val）从某种意义上说，限制是有限的，因为它无法从位置64向上预设位（因为unsigned long long是64位值）。高于64的任何位都将使用 0 进行初始化，如以下示例所示（参见位集构造函数）：

Bitset constructor constexpr bitset( unsigned long long val ) is limited in the sense that it cannot pre-set bits from position 64 upwards (as unsigned long long is a 64 bit value). Any bit higher than 64 will be initialized with 0, as the following example shows (cf. bitset constructors):

std::bitset<70> bl(ULLONG_MAX); // [0,0,0,0,0,0,1,1,1,...,1,1,1] in C++11

如果您还要初始化构造函数中的较高位，我想您就不会使用 string 或 char * ，即类型（3）或（4）的位集构造函数。

If you want to initialize also the higher bits in the constructor, I think you won't get around to use an initializer of type string or char*, i.e. bitset constructors of type (3) or (4).

当然，对于长度为256的位集，初始化程序字符串会变长：

Of course, for a bitset of length 256, the initializer string gets long:

std::bitset<256> myBitset("111100000011111000001110000010101000100000100010010000100000000001111");
// I did not count the digits...

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