首页 > 解决方案 > 为什么 __builtin_popcount 比我自己的位计数功能慢?

问题描述

在我编写了自己的位计数例程之后,我偶然发现了 gcc 的 __builtin_popcount。但是当我切换到 __builtin_popcount 时,我的软件实际上运行得更慢了。我在 Intel Core i3-4130T CPU @ 2.90GHz 上使用 Unbutu。我建立了一个性能测试来看看给出了什么。它看起来像这样:

#include <iostream>
#include <sys/time.h>
#include <stdint.h>

using namespace std;

const int bitCount[256] = {
    0,1,1,2,1,2,2,3,  1,2,2,3,2,3,3,4,  1,2,2,3,2,3,3,4,  2,3,3,4,3,4,4,5,
    1,2,2,3,2,3,3,4,  2,3,3,4,3,4,4,5,  2,3,3,4,3,4,4,5,  3,4,4,5,4,5,5,6,
    1,2,2,3,2,3,3,4,  2,3,3,4,3,4,4,5,  2,3,3,4,3,4,4,5,  3,4,4,5,4,5,5,6,
    2,3,3,4,3,4,4,5,  3,4,4,5,4,5,5,6,  3,4,4,5,4,5,5,6,  4,5,5,6,5,6,6,7,
    1,2,2,3,2,3,3,4,  2,3,3,4,3,4,4,5,  2,3,3,4,3,4,4,5,  3,4,4,5,4,5,5,6,
    2,3,3,4,3,4,4,5,  3,4,4,5,4,5,5,6,  3,4,4,5,4,5,5,6,  4,5,5,6,5,6,6,7,
    2,3,3,4,3,4,4,5,  3,4,4,5,4,5,5,6,  3,4,4,5,4,5,5,6,  4,5,5,6,5,6,6,7,
    3,4,4,5,4,5,5,6,  4,5,5,6,5,6,6,7,  4,5,5,6,5,6,6,7,  5,6,6,7,6,7,7,8
};

const uint32_t m32_0001 = 0x000000ffu;
const uint32_t m32_0010 = 0x0000ff00u;
const uint32_t m32_0100 = 0x00ff0000u;
const uint32_t m32_1000 = 0xff000000u;

inline int countBits(uint32_t bitField)
{
    return
        bitCount[(bitField & m32_0001)      ] +
        bitCount[(bitField & m32_0010) >>  8] +
        bitCount[(bitField & m32_0100) >> 16] +
        bitCount[(bitField & m32_1000) >> 24];
}

inline long long currentTime() {
    struct timeval ct;
    gettimeofday(&ct, NULL);
    return ct.tv_sec * 1000000LL + ct.tv_usec;
}

int main() {
    long long start, delta, sum;

    start = currentTime();
    sum = 0;
    for(unsigned i = 0; i < 100000000; ++i)
        sum += countBits(i);
    delta = currentTime() - start;
    cout << "countBits         : sum=" << sum << ": time (usec)=" << delta << endl;

    start = currentTime();
    sum = 0;
    for(unsigned i = 0; i < 100000000; ++i)
        sum += __builtin_popcount(i);
    delta = currentTime() - start;
    cout << "__builtin_popcount: sum=" << sum << ": time (usec)=" << delta << endl;

    start = currentTime();
    sum = 0;
    for(unsigned i = 0; i < 100000000; ++i) {
        int count;
        asm("popcnt %1,%0" : "=r"(count) : "rm"(i) : "cc");
        sum += count;
    }
    delta = currentTime() - start;
    cout << "assembler         : sum=" << sum << ": time (usec)=" << delta << endl;

    return 0;
}

起初,我使用较旧的编译器运行它:

> g++ --version | head -1
g++ (Ubuntu 4.8.4-2ubuntu1~14.04.3) 4.8.4
> cat /proc/cpuinfo | grep 'model name' | head -1
model name      : Intel(R) Core(TM) i3-4130T CPU @ 2.90GHz
> g++ -O3 popcountTest.cpp
> ./a.out
countBits         : sum=1314447104: time (usec)=148506
__builtin_popcount: sum=1314447104: time (usec)=345122
assembler         : sum=1314447104: time (usec)=138036

如您所见,基于表的 countBits 几乎与汇编程序一样快,并且远快于 __builtin_popcount。然后我在不同的机器类型上尝试了一个更新的编译器(相同的处理器——我认为主板也是一样的):

> g++ --version | head -1
g++ (Ubuntu 7.3.0-16ubuntu3) 7.3.0
> cat /proc/cpuinfo | grep 'model name' | head -1
model name      : Intel(R) Core(TM) i3-4130T CPU @ 2.90GHz
> g++ -O3 popcountTest.cpp
> ./a.out
countBits         : sum=1314447104: time (usec)=164247
__builtin_popcount: sum=1314447104: time (usec)=345167
assembler         : sum=1314447104: time (usec)=138028

奇怪的是,旧的编译器比新的编译器优化了我的 countBits 函数,但它仍然比汇编器好。显然,由于汇编程序行编译并运行,我的处理器支持 popcount,但为什么 __builtin_popcount 慢两倍以上?我自己的例程怎么可能与基于硅的 popcount 竞争?我在寻找第一个设置位等其他例程方面有相同的经验。我的例程都比 GNU“内置”等效项快得多。

(顺便说一句,我不知道如何编写汇编程序。我只是在某个网页上发现了那行,它似乎奇迹般地工作。)

标签: c++gcc

解决方案


如果没有在命令行上指定适当的“-march”,gcc 会生成对__popcountdi2函数而不是popcnt指令的调用。见:https ://godbolt.org/z/z1BihM

根据维基百科,自 Nehalem 以来英特尔和 AMD 自巴塞罗那以来都支持 POPCNT:https ://en.wikipedia.org/wiki/SSE4#POPCNT_and_LZCNT


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