什么更有效?使用 pow 平方或只是乘以它自己?

• With GCC 10 (no -ffast-math), x*x*x... is always faster
• With GCC 10 -O2 -ffast-math, std::pow is as fast as x*x*x... for odd exponents
• With GCC 10 -O3 -ffast-math, std::pow is as fast as x*x*x... for all test cases, and is around twice as fast as -O2.
• With GCC 10, C's pow(double, double) is always much slower
• With Clang 12 (no -ffast-math), x*x*x... is faster for exponents greater than 2
• With Clang 12 -ffast-math, all methods produce similar results
• With Clang 12, pow(double, double) is as fast as std::pow for integral exponents
• Writing benchmarks without having the compiler outsmart you is hard.

我最终会在我的机器上安装更新版本的 GCC，并在我这样做时更新我的​​结果.

I'll eventually get around to installing a more recent version of GCC on my machine and will update my results when I do so.

这是更新的基准代码:

#include <cmath>
#include <chrono>
#include <iostream>
#include <random>

using Moment = std::chrono::high_resolution_clock::time_point;
using FloatSecs = std::chrono::duration<double>;

inline Moment now()
{
    return std::chrono::high_resolution_clock::now();
}

#define TEST(num, expression) 
double test##num(double b, long loops) 
{ 
    double x = 0.0; 

    auto startTime = now(); 
    for (long i=0; i<loops; ++i) 
    { 
        x += expression; 
        b += 1.0; 
    } 
    auto elapsed = now() - startTime; 
    auto seconds = std::chrono::duration_cast<FloatSecs>(elapsed); 
    std::cout << seconds.count() << "	"; 
    return x; 
}

TEST(2, b*b)
TEST(3, b*b*b)
TEST(4, b*b*b*b)
TEST(5, b*b*b*b*b)

template <int exponent>
double testCppPow(double base, long loops)
{
    double x = 0.0;

    auto startTime = now();
    for (long i=0; i<loops; ++i)
    {
        x += std::pow(base, exponent);
        base += 1.0;
    }
    auto elapsed = now() - startTime;

    auto seconds = std::chrono::duration_cast<FloatSecs>(elapsed); 
    std::cout << seconds.count() << "	"; 

    return x;
}

double testCPow(double base, double exponent, long loops)
{
    double x = 0.0;

    auto startTime = now();
    for (long i=0; i<loops; ++i)
    {
        x += ::pow(base, exponent);
        base += 1.0;
    }
    auto elapsed = now() - startTime;

    auto seconds = std::chrono::duration_cast<FloatSecs>(elapsed); 
    std::cout << seconds.count() << "	"; 

    return x;
}

int main()
{
    using std::cout;
    long loops = 100000000l;
    double x = 0;
    std::random_device rd;
    std::default_random_engine re(rd());
    std::uniform_real_distribution<double> dist(1.1, 1.2);
    cout << "exp	c++ pow	c pow	x*x*x...";

    cout << "
2	";
    double b = dist(re);
    x += testCppPow<2>(b, loops);
    x += testCPow(b, 2.0, loops);
    x += test2(b, loops);

    cout << "
3	";
    b = dist(re);
    x += testCppPow<3>(b, loops);
    x += testCPow(b, 3.0, loops);
    x += test3(b, loops);

    cout << "
4	";
    b = dist(re);
    x += testCppPow<4>(b, loops);
    x += testCPow(b, 4.0, loops);
    x += test4(b, loops);

    cout << "
5	";
    b = dist(re);
    x += testCppPow<5>(b, loops);
    x += testCPow(b, 5.0, loops);
    x += test5(b, loops);

    std::cout << "
" << x << "
";
}

旧答案，2010 年

我使用此代码测试了 x*x*... 与 pow(x,i) 对于小型 i 之间的性能差异:

I tested the performance difference between x*x*... vs pow(x,i) for small i using this code:

#include <cstdlib>
#include <cmath>
#include <boost/date_time/posix_time/posix_time.hpp>

inline boost::posix_time::ptime now()
{
    return boost::posix_time::microsec_clock::local_time();
}

#define TEST(num, expression) 
double test##num(double b, long loops) 
{ 
    double x = 0.0; 

    boost::posix_time::ptime startTime = now(); 
    for (long i=0; i<loops; ++i) 
    { 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
        x += expression; 
    } 
    boost::posix_time::time_duration elapsed = now() - startTime; 

    std::cout << elapsed << " "; 

    return x; 
}

TEST(1, b)
TEST(2, b*b)
TEST(3, b*b*b)
TEST(4, b*b*b*b)
TEST(5, b*b*b*b*b)

template <int exponent>
double testpow(double base, long loops)
{
    double x = 0.0;

    boost::posix_time::ptime startTime = now();
    for (long i=0; i<loops; ++i)
    {
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
        x += std::pow(base, exponent);
    }
    boost::posix_time::time_duration elapsed = now() - startTime;

    std::cout << elapsed << " ";

    return x;
}

int main()
{
    using std::cout;
    long loops = 100000000l;
    double x = 0.0;
    cout << "1 ";
    x += testpow<1>(rand(), loops);
    x += test1(rand(), loops);

    cout << "
2 ";
    x += testpow<2>(rand(), loops);
    x += test2(rand(), loops);

    cout << "
3 ";
    x += testpow<3>(rand(), loops);
    x += test3(rand(), loops);

    cout << "
4 ";
    x += testpow<4>(rand(), loops);
    x += test4(rand(), loops);

    cout << "
5 ";
    x += testpow<5>(rand(), loops);
    x += test5(rand(), loops);
    cout << "
" << x << "
";
}

结果是:

1 00:00:01.126008 00:00:01.128338 
2 00:00:01.125832 00:00:01.127227 
3 00:00:01.125563 00:00:01.126590 
4 00:00:01.126289 00:00:01.126086 
5 00:00:01.126570 00:00:01.125930 
2.45829e+54

请注意，我累积了每次 pow 计算的结果，以确保编译器不会对其进行优化.

Note that I accumulate the result of every pow calculation to make sure the compiler doesn't optimize it away.

如果我使用 std::pow(double, double) 版本，并且 loops = 1000000l，我得到:

If I use the std::pow(double, double) version, and loops = 1000000l, I get:

1 00:00:00.011339 00:00:00.011262 
2 00:00:00.011259 00:00:00.011254 
3 00:00:00.975658 00:00:00.011254 
4 00:00:00.976427 00:00:00.011254 
5 00:00:00.973029 00:00:00.011254 
2.45829e+52

这是在运行 Ubuntu 9.10 64 位的 Intel Core Duo 上.使用带有 -o2 优化的 gcc 4.4.1 编译.

This is on an Intel Core Duo running Ubuntu 9.10 64bit. Compiled using gcc 4.4.1 with -o2 optimization.

所以在 C 中，是的 x*x*x 会比 pow(x, 3) 快，因为没有 pow(double, int) 重载.在 C++ 中，它大致相同.(假设我的测试方法是正确的.)

So in C, yes x*x*x will be faster than pow(x, 3), because there is no pow(double, int) overload. In C++, it will be the roughly same. (Assuming the methodology in my testing is correct.)

这是对 An Markm 的评论的回应:

This is in response to the comment made by An Markm:

即使发出了 using namespace std 指令，如果 pow 的第二个参数是 int，那么 std::pow(double, int) 来自 <cmath> 的重载将被调用，而不是来自 < 的 ::pow(double, double);math.h>.

Even if a using namespace std directive was issued, if the second parameter to pow is an int, then the std::pow(double, int) overload from <cmath> will be called instead of ::pow(double, double) from <math.h>.

此测试代码确认了该行为:

This test code confirms that behavior:

#include <iostream>

namespace foo
{

    double bar(double x, int i)
    {
        std::cout << "foo::bar
";
        return x*i;
    }


}

double bar(double x, double y)
{
    std::cout << "::bar
";
    return x*y;
}

using namespace foo;

int main()
{
    double a = bar(1.2, 3); // Prints "foo::bar"
    std::cout << a << "
";
    return 0;
}

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