从整数数组构造bitset [英] Construct bitset from array of integers

问题描述

很容易从 uint64_t 构建 bitset <64> ：

uint64_t flags = ...;
std::bitset<64> bs{flags};

但是有一个很好的方法来构造 bitset< 64 * N& uint64_t [N] 中的，使 flags [0] 最低64位？

But is there a good way to construct a bitset<64 * N> from a uint64_t[N], such that flags[0] would refer to the lowest 64 bits?

uint64_t flags[3];
// ... some assignments
std::bitset<192> bs{flags};  // this very unhelpfully compiles
                             // yet is totally invalid

在循环中调用 set（）？

推荐答案

std :: bitset 没有范围构造函数，所以你必须循环，但使用 std :: bitset :: set（）是underkill。 std :: bitset 支持二进制运算符，因此您至少可以批量设置64位：

std::bitset has no range constructor, so you will have to loop, but setting every bit individually with std::bitset::set() is underkill. std::bitset has support for binary operators, so you can at least set 64 bits in bulk:

  std::bitset<192> bs;

  for(int i = 2; i >= 0; --i) {
    bs <<= 64;
    bs |= flags[i];
  }

更新：在评论中，@icando比特转换对 std :: bitset 的O（N）操作的有效关注。对于非常大的比特组，这将最终吃掉大量处理的性能提升。在我的基准中，与单独设置比特并且不改变输入的简单循环相比， std :: bitset 的盈亏平衡点data：

Update: In the comments, @icando raises the valid concern that bitshifts are O(N) operations for std::bitsets. For very large bitsets, this will ultimately eat the performance boost of bulk processing. In my benchmarks, the break-even point for a std::bitset<N * 64> in comparison to a simple loop that sets the bits individually and does not mutate the input data:

int pos = 0;
for(auto f : flags) {
  for(int b = 0; b < 64; ++b) {
    bs.set(pos++, f >> b & 1);
  }
}

= 200 （使用libstdc ++和 -O2 在x86-64上为gcc 4.9）。 Clang执行有点差，甚至破坏 N == 160 。 Gcc与 -O3 将其推向 N == 250 。

is somewhere around N == 200 (gcc 4.9 on x86-64 with libstdc++ and -O2). Clang performs somewhat worse, breaking even around N == 160. Gcc with -O3 pushes it up to N == 250.

取较低端，这意味着如果要使用10000位或更大的位组，这种方法可能不适合您。在32位平台（如常见ARM ），阈值可能会降低，因此在这种平台上使用5000位位组时请记住。然而，我认为，在这之前的某个地方，你应该问自己，一个bitset是否真的是容器的正确选择。

Taking the lower end, this means that if you want to work with bitsets of 10000 bits or larger, this approach may not be for you. On 32-bit platforms (such as common ARMs), the threshold will probably lie lower, so keep that in mind when you work with 5000-bit bitsets on such platforms. I would argue, however, that somewhere far before this point, you should have asked yourself if a bitset is really the right choice of container.

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