何时在`std :: sqrt（x * x + y * y）上使用`std :: hypot（x，y）`

std :: hypot 文档表示：

计算x和y平方和的平方根，中间不会出现不适当的上溢或下溢

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.

我很难设想一个测试用例，其中 std :: hypot 应该用于普通的 sqrt（x * x + y * y）。

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

以下测试表明 std :: hypot 大约比朴素的计算慢20倍。

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;
    }
}

所以我要寻求指导，我什么时候必须使用 std :: hypot（x，y）在更快的 std :: sqrt（x * x + y * y）上获得正确的结果。

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).

说明：我正在寻找 x时适用的答案和 y 是浮点数。即比较：

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

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);