Idiom #200 Return hypotenuse

Compute the hypotenuse h of the triangle where the sides adjacent to the square angle have lengths x and y.

const h = Math.hypot(x, y);

var h = Math.sqrt(x*x + y*y);

#include <cmath>

auto h = std::hypot(x, y);

double hypo(double x, double y)
{
    return Math.Sqrt(Math.Pow(x, 2) + Math.Pow(y, 2));
}

import 'dart:math';

var h = sqrt(x * x + y * y);

import :math

def sq(x) do
  x*x
end

def hypo(a,b) do
  sqrt(sq(a) + sq(b))
end

h = hypot(x,y)

import "math"

h := math.Hypot(x, y)

hypo x y = sqrt $ x**2 + y**2

import static java.lang.Math.hypot;

double h = hypot(x, y);

return double h = Math.sqrt(Math.pow(x,2)+Math.pow(y,2));

import static java.math.MathContext.DECIMAL128;
import java.math.BigDecimal;

BigDecimal X = new BigDecimal(x).pow(2),
           Y = new BigDecimal(y).pow(2),
           h = X.add(Y).sqrt(DECIMAL128);

import static java.lang.Math.sqrt;

double h = sqrt((x * x) + (y * y));

local h = math.sqrt(x^2 + y^2)

local h=(x^2+y^2)^0.5

$h = hypot($x, $y)

// before PHP 4.1
$h = sqrt($x*$x + $y*$y);

uses math;

h := hypot(x,y);

use Math::AnyNum qw(hypot);

my $h = hypot $x, $y;

import math

h = math.hypot(x, y)

include Math

h = hypot(x, y)

let h = x.hypot(y);

fn hypot(x:f64, y:f64)-> f64 {
    let num = x.powi(2) + y.powi(2);
    num.powf(0.5)
}

Do you know the best way to do this in your language ? New implementation...

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Idiom created by Bart

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