Easiest Way to Find Square Root in Swift

Best way to calculate the square root of any number in ios , Objective C and Swift

The sqrt function (and other mathematical functions as well) is
available in the standard libraries on all OS X and iOS platforms.

It can be used from (Objective-)C:

#include "math.h"

double sqrtFive = sqrt(5.0);

and from Swift:

import Darwin // or Foundation, Cocoa, UIKit, ...

let sqrtFive = sqrt(5.0)

Swift 5 calling squareRoot()

squareRoot() is a method defined in the FloatingPoint protocol, to which Double and Float conform, but not integer types like Int.

Therefore you have to convert the Int to Double (and the result back to Int):

let upperBound = Int(Double(number).squareRoot())
for i in 2..<upperBound {

The other example

let test = 100.squareRoot()

compiles because both Int and Double conform to the ExpressibleByIntegerLiteral protocol, i.e. can be initialized from an integer literal. Here the compiler automatically infers 100 to be a floating point value because that is the only way to make the code compile. It becomes obvious here:

let test = 123.squareRoot()
print(test) // 11.090536506409418

Generic square root in Swift

I see that you're using using a custom Arithmetic protocol to constraint the generic.

My approach would be to declare 2 required methods in that protocol: toDouble() and fromDouble(), and implement both in Float, Double and Int extensions. Note that fromDouble() should be a static method.

This way you can convert T to Double, hence be able to use sqrt(), and convert back from Double to T.

Last, there's a bug in your code: if left is an empty vector, the function will crash, because the code in the loop will never be executed, so result will keep its nil initial value. The forced unwrapping in the return statement will fail, causing the exception.

Rooting without using sqrt in Swift

OK, so here's your code after I aligned it.

import Foundation

func rooting(number: Int) -> Int {
  for i in 1...100 {
    if i * i == number {
      var Root = i
      break
    }
    return Root
  }
  return Root
}

print(rooting(number: 9))

The problem is that you are creating the Root variable in a scope that ends before you try to use the variable. Essentially, the Root variable only exists in the following segment:

import Foundation

func rooting(number: Int) -> Int {
  for i in 1...100 {
    if i * i == number {
      var Root = I //Root variable starts existing
      break
    } //Root variable no longer exists
    return Root
  }
  return Root
}

print(rooting(number: 9))

This is because it is scoped to the if statement. To solve this specific problem, you could change your code to this:

import Foundation

func rooting(number: Int) -> Int {
  for i in 1...100 {
    if i * i == number {
      return i
    }
  }
}

print(rooting(number: 9))

since you don't do anything with Root other than return it. Of course, this still won't work, as you aren't guaranteed to return anything (what if number is 8?), and you said that the function will always return an Int.

---

Here's a quick example of an implementation of a rooting function:

import Foundation

func rooting(number: Int) -> Int? {
  guard number > 0 else { return nil }
  for i in 0...(number/2) {
    if i * i == number {
      return i
    }
  }
  return nil
}

print(rooting(number: 9))

If the number has a square root, it will be returned; otherwise, the function returns nil.

How to calculate the 6th root of 2 in Swift?

To compute the 6th root of 2:

pow(2.0, 1.0/6.0)

You just need to use Doubles.

There are multiple versions of pow for different data types. By using the floating point literals, Swift will infer them be of type Double and select pow with this signature:

func pow(_: Double, _: Double) -> Double

Or you can use logarithms (nth root of x is exp(log(x)/n) where x and n are Doubles:

exp(log(2.0) / 6.0)

how to get the nearest Int floored from a sqrt of an Int?

If you want to call generatePrimes with a Double

generatePrimes(to: floor(sqrt(Double(n)))

or if you want to call it with an Int

generatePrimes(to: Int(floor(sqrt(Double(n))))

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