Math.hypot() - JavaScript 编辑

The Math.hypot() function returns the square root of the sum of squares of its arguments, that is:

$$\mathtt{\operatorname{Math.hypot}(v_1, v_2, \dots, v_n)} = \sqrt{\sum_{i=1}^n v_i^2} = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}$$

The source for this interactive example is stored in a GitHub repository. If you'd like to contribute to the interactive examples project, please clone https://github.com/mdn/interactive-examples and send us a pull request.

The source for this interactive example is stored in a GitHub repository. If you'd like to contribute to the interactive examples project, please clone https://github.com/mdn/interactive-examples and send us a pull request.

Syntax

Math.hypot([value1[, value2[, ...]]])

Parameters

value1, value2, ...

Numbers.

Return value

The square root of the sum of squares of the given arguments. If at least one of the arguments cannot be converted to a number, NaN is returned.

Description

Calculating the hypotenuse of a right triangle, or the magnitude of a complex number, uses the formula Math.sqrt(v1*v1 + v2*v2), where v1 and v2 are the lengths of the triangle's legs, or the complex number's real and complex components. The corresponding distance in 2 or more dimensions can be calculated by adding more squares under the square root: Math.sqrt(v1*v1 + v2*v2 + v3*v3 + v4*v4).

This function makes this calculation easier and faster; you call Math.hypot(v1, v2) , or Math.hypot(v1, v2, v3, v4, ...).

Math.hypot also avoids overflow/underflow problems if the magnitude of your numbers is very large. The largest number you can represent in JS is Number.MAX_VALUE, which is around 10³⁰⁸. If your numbers are larger than about 10¹⁵⁴, taking the square of them will result in Infinity. For example, Math.sqrt(1e200*1e200 + 1e200*1e200) = Infinity. If you use hypot() instead, you get better answer: Math.hypot(1e200, 1e200) = 1.4142...e+200 . This is also true with very small numbers. Math.sqrt(1e-200*1e-200 + 1e-200*1e-200) = 0, but Math.hypot(1e-200, 1e-200) = 1.4142...e-200.

Because hypot() is a static method of Math, you always use it as Math.hypot(), rather than as a method of a Math object you created (Math is not a constructor).

If no arguments are given, the result is +0. If any of the arguments is ±Infinity, the result is Infinity. If any of the arguments is NaN (unless another argument is ±Infinity), the result is NaN. If at least one of the arguments cannot be converted to a number, the result is NaN.

With one argument, Math.hypot() is equivalent to Math.abs().

Examples

Using Math.hypot()

Math.hypot(3, 4);          // 5
Math.hypot(3, 4, 5);       // 7.0710678118654755
Math.hypot();              // 0
Math.hypot(NaN);           // NaN
Math.hypot(NaN, Infinity); // Infinity
Math.hypot(3, 4, 'foo');   // NaN, since +'foo' => NaN
Math.hypot(3, 4, '5');     // 7.0710678118654755, +'5' => 5
Math.hypot(-3);            // 3, the same as Math.abs(-3)

Polyfill

A naive approach that does not handle overflow/underflow issues:

if (!Math.hypot) Math.hypot = function() {
  var y = 0, i = arguments.length, containsInfinity = false;
  while (i--) {
    var arg = arguments[i];
    if (arg === Infinity || arg === -Infinity)
      containsInfinity = true
    y += arg * arg
  }
  return containsInfinity ? Infinity : Math.sqrt(y)
}

A polyfill that avoids underflows and overflows:

if (!Math.hypot) Math.hypot = function () {
  var max = 0;
  var s = 0;
  var containsInfinity = false;
  for (var i = 0; i < arguments.length; ++i) {
    var arg = Math.abs(Number(arguments[i]));
    if (arg === Infinity)
      containsInfinity = true
    if (arg > max) {
      s *= (max / arg) * (max / arg);
      max = arg;
    }
    s += arg === 0 && max === 0 ? 0 : (arg / max) * (arg / max);
  }
  return containsInfinity ? Infinity : (max === 1 / 0 ? 1 / 0 : max * Math.sqrt(s));
};

Specifications

Specification
ECMAScript (ECMA-262) The definition of 'Math.hypot' in that specification.

Browser compatibility

BCD tables only load in the browser

See also

• Math.abs()
• Math.pow()
• Math.sqrt()