何时使用`std :: hypot（x，y）`而不是`std :: sqrt（x * x + y * y）`

Question

The documentation of std::hypot says that:

Computes the square root of the sum of the squares of x and y, without undue overflow or underflow at intermediate stages of the computation.

I struggle to conceive a test case where std::hypot should be used over the trivial sqrt(x*x + y*y) .

The following test shows that std::hypot is roughly 20x slower than the naive calculation.

#include <iostream>
#include <chrono>
#include <random>
#include <algorithm>

int main(int, char**) {
    std::mt19937_64 mt;
    const auto samples = 10000000;
    std::vector<double> values(2 * samples);
    std::uniform_real_distribution<double> urd(-100.0, 100.0);
    std::generate_n(values.begin(), 2 * samples, [&]() {return urd(mt); });
    std::cout.precision(15);

    {
        double sum = 0;
        auto s = std::chrono::steady_clock::now();
        for (auto i = 0; i < 2 * samples; i += 2) {
            sum += std::hypot(values[i], values[i + 1]);
        }
        auto e = std::chrono::steady_clock::now();
        std::cout << std::fixed <<std::chrono::duration_cast<std::chrono::microseconds>(e - s).count() << "us --- s:" << sum << std::endl;
    }
    {
        double sum = 0;
        auto s = std::chrono::steady_clock::now();
        for (auto i = 0; i < 2 * samples; i += 2) {
            sum += std::sqrt(values[i]* values[i] + values[i + 1]* values[i + 1]);
        }
        auto e = std::chrono::steady_clock::now();
        std::cout << std::fixed << std::chrono::duration_cast<std::chrono::microseconds>(e - s).count() << "us --- s:" << sum << std::endl;
    }
}

So I'm asking for guidance, when must I use std::hypot(x,y) to obtain correct results over the much faster std::sqrt(x*x + y*y) .

Clarification: I'm looking for answers that apply when x and y are floating point numbers. Ie compare:

double h = std::hypot(static_cast<double>(x),static_cast<double>(y));

to:

double xx = static_cast<double>(x);
double yy = static_cast<double>(y);
double h = std::sqrt(xx*xx + yy*yy);

Answer 1

The answer is in the documentation you quoted

Computes the square root of the sum of the squares of x and y, without undue overflow or underflow at intermediate stages of the computation .

If x*x + y*y overflows, then if you carry out the calculation manually, you'll get the wrong answer. If you use std::hypot , however, it guarantees that the intermediate calculations will not overflow.

You can see an example of this disparity here .

If you are working with numbers which you know will not overflow the relevant representation for your platform, you can happily use the naive version.