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》
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