c – 什么时候在`std :: sqrt(x * x y * y)’上使用`std :: hypo

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.

我很难想象一个测试用例,其中std :: hypot应该用于平凡的sqrt(x * x y * y).

以下测试显示,std :: hypot比原始计算慢约20倍.

#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;
    }
}

所以我要求指导,什么时候我必须使用std :: hypot(x,y)通过更快的std :: sqrt(x * x y * y)获得正确的结果.

澄清：当x和y是浮点数时,我正在寻找适用的答案.即比较：

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

至：

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