二进制中 1 的个数 ——《C/C++ 位运算黑科技 03》

原理

计算一个二进制数中 1 的出现次数其实很简单，只需要不断用 v & (v - 1) 移除掉最后一个 1 即可，原理可以参考这篇文章：2 的幂次方 ——《C/C++ 位运算黑科技 02》在新窗口打开

上述方法是一个普通的思考方向，下面我会介绍另外一种思路：并行计数器，来计算二进制数中出现的 1

实际上，我们可以将这个数看作是全部由单位的计数器组成，1、0 就代表单个计数器的状态，我们只要合并相邻的计数器即可，这其实也是归并的思想。

代码

inline unsigned count_bits(uint64_t v)
{
    v = (v & 0x5555555555555555) + ((v >> 1) & 0x5555555555555555);
    v = (v & 0x3333333333333333) + ((v >> 2) & 0x3333333333333333);
    v = (v & 0x0f0f0f0f0f0f0f0f) + ((v >> 4) & 0x0f0f0f0f0f0f0f0f);
    v = (v & 0x00ff00ff00ff00ff) + ((v >> 8) & 0x00ff00ff00ff00ff);
    v = (v & 0x0000ffff0000ffff) + ((v >> 16) & 0x0000ffff0000ffff);
    v = (v & 0x00000000ffffffff) + ((v >> 32) & 0x00000000ffffffff);
    return v;
}

inline unsigned count_bits(uint32_t v)
{
    v = (v & 0x55555555) + ((v >> 1) & 0x55555555);
    v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
    v = (v & 0x0f0f0f0f) + ((v >> 4) & 0x0f0f0f0f);
    v = (v & 0x00ff00ff) + ((v >> 8) & 0x00ff00ff);
    v = (v & 0x0000ffff) + ((v >> 16) & 0x0000ffff);
    return v;
}

inline unsigned count_bits(uint16_t v)
{
    v = (v & 0x5555) + ((v >> 1) & 0x5555);
    v = (v & 0x3333) + ((v >> 2) & 0x3333);
    v = (v & 0x0f0f) + ((v >> 4) & 0x0f0f);
    v = (v & 0x00ff) + ((v >> 8) & 0x00ff);
    return v;
}

inline unsigned count_bits(uint8_t v)
{
    v = (v & 0x55) + ((v >> 1) & 0x55);
    v = (v & 0x33) + ((v >> 2) & 0x33);
    v = (v & 0x0f) + ((v >> 4) & 0x0f);
    return v;
}

原理剖析

下面以 1110001010011110 作为例子，来解释并行计数器合并的方法：

val	1	1	1	0	0	0	1	0	1	0	0	1	1	1	1	0
& 0x5555	0	1	0	1	0	1	0	1	0	1	0	1	0	1	0	1
=	0	1	0	0	0	0	0	0	0	0	0	1	0	1	0	0

val >> 1	0	1	1	1	0	0	0	1	0	1	0	0	1	1	1	1
& 0x5555	0	1	0	1	0	1	0	1	0	1	0	1	0	1	0	1
=	0	1	0	1	0	0	0	1	0	1	0	0	0	1	0	1

然后两者相加就得到了相邻 2 个计数器的合并计数：1001000101011001，然后我们在以 2 个比特为单位来继续合并计数器

Val	10	01	00	01	01	01	10	01
& 0x3333	00	11	00	11	00	11	00	11
=	00	01	00	01	00	01	00	01

Val >> 2	00	10	01	00	01	01	01	10
& 0x3333	00	11	00	11	00	11	00	11
=	00	10	00	00	00	01	00	10

然后两者相加就得到了相邻 4 个计数器的合并计数：0011000100100011，然后我们在以 4 个比特为单位来继续合并计数器

Val	0011	0001	0010	0011
& 0x0f0f	0000	1111	0000	1111
=	0000	0001	0000	0011

Val >> 4	0000	0011	0001	0010
& 0x0f0f	0000	1111	0000	1111
=	0000	0011	0000	0010

然后两者相加就得到了相邻 8 个计数器的合并计数：0000010000000101，然后我们在以 8 个比特为单位来继续合并计数器

Val	00000100	00000101
&00ff	00000000	11111111
=	00000000	00000101

Val >> 8	00000000	00000100
&00ff	00000000	11111111
=	00000000	00000100

然后两者相加就得到了相邻 8 个计数器的合并计数：0000000000001001，转换成十进制就是 9，与原数字中的 1 的个数是相同的。

Benchmark

#include "benchmark/benchmark.h"

inline unsigned count_bits(uint64_t v)
{
  v = (v & 0x5555555555555555) + ((v >> 1) & 0x5555555555555555);
  v = (v & 0x3333333333333333) + ((v >> 2) & 0x3333333333333333);
  v = (v & 0x0f0f0f0f0f0f0f0f) + ((v >> 4) & 0x0f0f0f0f0f0f0f0f);
  v = (v & 0x00ff00ff00ff00ff) + ((v >> 8) & 0x00ff00ff00ff00ff);
  v = (v & 0x0000ffff0000ffff) + ((v >> 16) & 0x0000ffff0000ffff);
  v = (v & 0x00000000ffffffff) + ((v >> 32) & 0x00000000ffffffff);
  return v;
}

inline unsigned count_bits(uint32_t v)
{
  v = (v & 0x55555555) + ((v >> 1) & 0x55555555);
  v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
  v = (v & 0x0f0f0f0f) + ((v >> 4) & 0x0f0f0f0f);
  v = (v & 0x00ff00ff) + ((v >> 8) & 0x00ff00ff);
  v = (v & 0x0000ffff) + ((v >> 16) & 0x0000ffff);
  return v;
}

inline unsigned count_bits(uint16_t v)
{
  v = (v & 0x5555) + ((v >> 1) & 0x5555);
  v = (v & 0x3333) + ((v >> 2) & 0x3333);
  v = (v & 0x0f0f) + ((v >> 4) & 0x0f0f);
  v = (v & 0x00ff) + ((v >> 8) & 0x00ff);
  return v;
}

inline unsigned count_bits(uint8_t v)
{
  v = (v & 0x55) + ((v >> 1) & 0x55);
  v = (v & 0x33) + ((v >> 2) & 0x33);
  v = (v & 0x0f) + ((v >> 4) & 0x0f);
  return v;
}

static void BM_count_64(benchmark::State &state) {
  for (auto _: state) {
    uint64_t n = UINT64_MAX;
    benchmark::DoNotOptimize(count_bits(n));
  }
}

static void BM_count_32(benchmark::State &state) {
  for (auto _: state) {
    uint32_t n = UINT32_MAX;
    benchmark::DoNotOptimize(count_bits(n));
  }
}

static void BM_count_16(benchmark::State &state) {
  for (auto _: state) {
    uint16_t n = UINT16_MAX;
    benchmark::DoNotOptimize(count_bits(n));
  }
}

static void BM_count_8(benchmark::State &state) {
  for (auto _: state) {
    uint8_t n = UINT8_MAX;
    benchmark::DoNotOptimize(count_bits(n));
  }
}

BENCHMARK(BM_count_8);
BENCHMARK(BM_count_16);
BENCHMARK(BM_count_32);
BENCHMARK(BM_count_64);

BENCHMARK_MAIN();

下面是使用 MacBook Air (M1, 2020) 和 Apple clang 13.1.6 得到的结果

/Users/hominsu/CLionProjects/bit-hacks-bench/cmake-build-release-appleclang/bench/count_bits
Unable to determine clock rate from sysctl: hw.cpufrequency: No such file or directory
2022-03-27T14:09:30+08:00
Running /Users/hominsu/CLionProjects/bit-hacks-bench/cmake-build-release-appleclang/bench/count_bits
Run on (8 X 24.1205 MHz CPU s)
CPU Caches:
  L1 Data 64 KiB (x8)
  L1 Instruction 128 KiB (x8)
  L2 Unified 4096 KiB (x2)
Load Average: 2.64, 2.22, 1.79
------------------------------------------------------
Benchmark            Time             CPU   Iterations
------------------------------------------------------
BM_count_8       0.319 ns        0.319 ns   1000000000
BM_count_16      0.321 ns        0.321 ns   1000000000
BM_count_32      0.313 ns        0.313 ns   1000000000
BM_count_64      0.316 ns        0.316 ns   1000000000

下面是使用 i5-9500 和 gcc 8.5.0 (Red Hat 8.5.0-10) 在 CentOS-8-Stream 下得到的结果

/tmp/tmp.CtmwmpTLjC/cmake-build-release-1104/bench/count_bits
2022-03-27T14:10:07+08:00
Running /tmp/tmp.CtmwmpTLjC/cmake-build-release-1104/bench/count_bits
Run on (6 X 4100.35 MHz CPU s)
CPU Caches:
  L1 Data 32 KiB (x6)
  L1 Instruction 32 KiB (x6)
  L2 Unified 256 KiB (x6)
  L3 Unified 9216 KiB (x1)
Load Average: 0.57, 0.54, 0.51
------------------------------------------------------
Benchmark            Time             CPU   Iterations
------------------------------------------------------
BM_count_8       0.244 ns        0.244 ns   1000000000
BM_count_16      0.246 ns        0.246 ns   1000000000
BM_count_32      0.245 ns        0.244 ns   1000000000
BM_count_64      0.249 ns        0.248 ns   1000000000

# 二进制中 1 的个数 ——《C/C++ 位运算黑科技 03》

# 原理

# 代码

# 原理剖析

# Benchmark

二进制中 1 的个数 ——《C/C++ 位运算黑科技 03》

原理

代码

原理剖析

Benchmark