What is more efficient? Using pow to square or just multiply it with itself?

Question

UPDATE 2021

I've modified the benchmark code as follows:

• std::chrono used for timing measurements instead of boost
• C++11 <random> used instead of rand()
• Avoid repeated operations that can get hoisted out. The base parameter is ever-changing.

I get the following results with GCC 10 -O2 (in seconds):

exp     c++ pow     c pow       x*x*x...
2       0.204243    1.39962     0.0902527   
3       1.36162     1.38291     0.107679    
4       1.37717     1.38197     0.106103    
5       1.3815      1.39139     0.117097

GCC 10 -O3 is almost identical to GCC 10 -O2.

With GCC 10 -O2 -ffast-math:

exp     c++ pow     c pow       x*x*x...
2       0.203625    1.4056      0.0913414   
3       0.11094     1.39938     0.108027    
4       0.201593    1.38618     0.101585    
5       0.102141    1.38212     0.10662

With GCC 10 -O3 -ffast-math:

exp     c++ pow     c pow       x*x*x...
2       0.0451995   1.175       0.0450497   
3       0.0470842   1.20226     0.051399    
4       0.0475239   1.18033     0.0473844   
5       0.0522424   1.16817     0.0522291

With Clang 12 -O2:

exp     c++ pow     c pow       x*x*x...
2       0.106242    0.105435    0.105533    
3       1.45909     1.4425      0.102235    
4       1.45629     1.44262     0.108861    
5       1.45837     1.44483     0.1116

Clang 12 -O3 is almost identical to Clang 12 -O2.

With Clang 12 -O2 -ffast-math:

exp     c++ pow     c pow       x*x*x...
2       0.0233731   0.0232457   0.0231076   
3       0.0271074   0.0266663   0.0278415   
4       0.026897    0.0270698   0.0268115   
5       0.0312481   0.0296402   0.029811    

Clang 12 -O3 -ffast-math is almost identical to Clang 12 -O2 -ffast-math.

Machine is Intel Core i7-7700K on Linux 5.4.0-73-generic x86_64.

Conclusions:

• 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.

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.

Here's the updated benchmark code:

#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() << "\t"; \
    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() << "\t"; \

    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() << "\t"; \

    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\tc++ pow\tc pow\tx*x*x...";

    cout << "\n2\t";
    double b = dist(re);
    x += testCppPow<2>(b, loops);
    x += testCPow(b, 2.0, loops);
    x += test2(b, loops);

    cout << "\n3\t";
    b = dist(re);
    x += testCppPow<3>(b, loops);
    x += testCPow(b, 3.0, loops);
    x += test3(b, loops);

    cout << "\n4\t";
    b = dist(re);
    x += testCppPow<4>(b, loops);
    x += testCPow(b, 4.0, loops);
    x += test4(b, loops);

    cout << "\n5\t";
    b = dist(re);
    x += testCppPow<5>(b, loops);
    x += testCPow(b, 5.0, loops);
    x += test5(b, loops);

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

---

Old Answer, 2010

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 << "\n2 ";
    x += testpow<2>(rand(), loops);
    x += test2(rand(), loops);

    cout << "\n3 ";
    x += testpow<3>(rand(), loops);
    x += test3(rand(), loops);

    cout << "\n4 ";
    x += testpow<4>(rand(), loops);
    x += test4(rand(), loops);

    cout << "\n5 ";
    x += testpow<5>(rand(), loops);
    x += test5(rand(), loops);
    cout << "\n" << x << "\n";
}

Results are:

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

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

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

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

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.)

---

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

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>.

This test code confirms that behavior:

#include <iostream>

namespace foo
{

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


}

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

using namespace foo;

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

That's the wrong kind of question. The right question would be: "Which one is easier to understand for human readers of my code?"

If speed matters (later), don't ask, but measure. (And before that, measure whether optimizing this actually will make any noticeable difference.) Until then, write the code so that it is easiest to read.

Edit
Just to make this clear (although it already should have been): Breakthrough speedups usually come from things like using better algorithms, improving locality of data, reducing the use of dynamic memory, pre-computing results, etc. They rarely ever come from micro-optimizing single function calls, and where they do, they do so in very few places, which would only be found by careful (and time-consuming) profiling, more often than never they can be sped up by doing very non-intuitive things (like inserting noop statements), and what's an optimization for one platform is sometimes a pessimization for another (which is why you need to measure, instead of asking, because we don't fully know/have your environment).

Let me underline this again: Even in the few applications where such things matter, they don't matter in most places they're used, and it is very unlikely that you will find the places where they matter by looking at the code. You really do need to identify the hot spots first, because otherwise optimizing code is just a waste of time.

Even if a single operation (like computing the square of some value) takes up 10% of the application's execution time (which IME is quite rare), and even if optimizing it saves 50% of the time necessary for that operation (which IME is even much, much rarer), you still made the application take only 5% less time.
Your users will need a stopwatch to even notice that. (I guess in most cases anything under 20% speedup goes unnoticed for most users. And that is four such spots you need to find.)

x*x or x*x*x will be faster than pow, since pow must deal with the general case, whereas x*x is specific. Also, you can elide the function call and suchlike.

However, if you find yourself micro-optimizing like this, you need to get a profiler and do some serious profiling. The overwhelming probability is that you would never notice any difference between the two.

I was also wondering about the performance issue, and was hoping this would be optimised out by the compiler, based on the answer from @EmileCormier. However, I was worried that the test code he showed would still allow the compiler to optimise away the std::pow() call, since the same values were used in the call every time, which would allow the compiler to store the results and re-use it in the loop - this would explain the almost identical run-times for all cases. So I had a look into it too.

Here's the code I used (test_pow.cpp):

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

class Timer {
  public:
    explicit Timer () : from (std::chrono::high_resolution_clock::now()) { }

    void start () {
      from = std::chrono::high_resolution_clock::now();
    }

    double elapsed() const {
      return std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::high_resolution_clock::now() - from).count() * 1.0e-6;
    }

  private:
    std::chrono::high_resolution_clock::time_point from;
};

int main (int argc, char* argv[])
{
  double total;
  Timer timer;



  total = 0.0;
  timer.start();
  for (double i = 0.0; i < 1.0; i += 1e-8)
    total += std::pow (i,2);
  std::cout << "std::pow(i,2): " << timer.elapsed() << "s (result = " << total << ")\n";

  total = 0.0;
  timer.start();
  for (double i = 0.0; i < 1.0; i += 1e-8)
    total += i*i;
  std::cout << "i*i: " << timer.elapsed() << "s (result = " << total << ")\n";

  std::cout << "\n";

  total = 0.0;
  timer.start();
  for (double i = 0.0; i < 1.0; i += 1e-8)
    total += std::pow (i,3);
  std::cout << "std::pow(i,3): " << timer.elapsed() << "s (result = " << total << ")\n";

  total = 0.0;
  timer.start();
  for (double i = 0.0; i < 1.0; i += 1e-8)
    total += i*i*i;
  std::cout << "i*i*i: " << timer.elapsed() << "s (result = " << total << ")\n";


  return 0;
}

This was compiled using:

g++ -std=c++11 [-O2] test_pow.cpp -o test_pow

Basically, the difference is the argument to std::pow() is the loop counter. As I feared, the difference in performance is pronounced. Without the -O2 flag, the results on my system (Arch Linux 64-bit, g++ 4.9.1, Intel i7-4930) were:

std::pow(i,2): 0.001105s (result = 3.33333e+07)
i*i: 0.000352s (result = 3.33333e+07)

std::pow(i,3): 0.006034s (result = 2.5e+07)
i*i*i: 0.000328s (result = 2.5e+07)

With optimisation, the results were equally striking:

std::pow(i,2): 0.000155s (result = 3.33333e+07)
i*i: 0.000106s (result = 3.33333e+07)

std::pow(i,3): 0.006066s (result = 2.5e+07)
i*i*i: 9.7e-05s (result = 2.5e+07)

So it looks like the compiler does at least try to optimise the std::pow(x,2) case, but not the std::pow(x,3) case (it takes ~40 times longer than the std::pow(x,2) case). In all cases, manual expansion performed better - but particularly for the power 3 case (60 times quicker). This is definitely worth bearing in mind if running std::pow() with integer powers greater than 2 in a tight loop...

The most efficient way is to consider the exponential growth of the multiplications. Check this code for p^q:

template <typename T>
T expt(T p, unsigned q){
    T r =1;
    while (q != 0) {
        if (q % 2 == 1) {    // if q is odd
            r *= p;
            q--;
        }
        p *= p;
        q /= 2;
    }
    return r;
}