Why is this handmade square root function in Swift not returning the value, but throws the error?-CodePudding

Just trying to create an analog of sqrt() in Swift, but it throws .noRoot in all the matched cases. Also I add the error .outOfBound, this is working correctly.

import UIKit

enum WrongNumber: Error {
  case outOfBounds
  case noRoot
}

func mySqrt(_ number: Int) throws -> Int {
  if number < 1 || number > 10_000 {
    throw WrongNumber.outOfBounds
  }
  
  var result = 0
    for i in 1...10_000 {
      if i * i == number {
        result = i
      } else {
        throw WrongNumber.noRoot
      }
   } 
  return result
}

var number = 1000

do {
  let result = try mySqrt(number)
    print(result)
}
catch WrongNumber.outOfBounds {
    print("You're puting a wrong number. Please try again with range from 1 to 10_000")
 }
catch WrongNumber.noRoot {
    print("There is no square root in your number")
}

CodePudding user response：

You should return once you find it

func mySqrt(_ number: Int) throws -> Int {
  if number < 1 || number > 10_000 {
    throw WrongNumber.outOfBounds
  }
  for i in 1...10_000 {
    if i * i == number {
      return i
    }
  } 
  throw WrongNumber.noRoot
}