for in loop with where clause in Swift

Question

I have tried to update a little function to Swift 2.1. The original working code was:

import func Darwin.sqrt
func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }

func sigma(n: Int) -> Int {
    // adding up proper divisors from 1 to sqrt(n) by trial divison
    if n == 1 { return 0 } // definition of aliquot sum
    var result = 1
    let root = sqrt(n)
    for var div = 2; div <= root; ++div {
        if n % div == 0 {
            result += div + n/div
        }
    }
    if root*root == n { result -= root }
    return (result)
}
print(sigma(10))
print(sigma(3))

After updating the for loop I get a runtime error for the last line. Any idea why that happens?

import func Darwin.sqrt
func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }

func sigma(n: Int) -> Int {
    // adding up proper divisors from 1 to sqrt(n) by trial divison
    if n == 1 { return 0 } // definition of aliquot sum
    var result = 1
    let root = sqrt(n)
    for div in 2...root where n % div == 0 {
            result += div + n/div
    }
    if root*root == n { result -= root }
    return (result)
}
print(sigma(10))
print(sigma(3)) //<- run time error with for in loop

Since you are updating your code, you might want to change if n == 1 {return 0} to guard n != 1 else { return 0 } — TPlet
– TPlet, Commented Jan 8, 2016 at 11:13
Using stride() is another option, compare stackoverflow.com/questions/34323227/…. — Martin R
– Martin R, Commented Jan 8, 2016 at 11:33
@ramkrishna-sharma Don't tag with "ios" questions that are not directly related to iOS. Swift is also supported by OS X and now Linux; Swift questions may apply and are not specific to iOS. In short: Swift ≠ iOS. Thank you. — Eric Aya
– Eric Aya, Commented Jan 8, 2016 at 11:36

Sergey Kalinichenko · Accepted Answer · 2016-01-08 11:09:20Z

6

When you pass 3 to sigma, your range 2...root becomes invalid, because the left side, the root, is less than the right side, 2.

The closed range operator (a...b) defines a range that runs from a to b, and includes the values a and b. The value of a must not be greater than b.

root is assigned sqrt(n), which means that in order for the 2...root range to remain valid, n must be above 2².

You can fix by supplying a lower limit for the right side, i.e.

for div in 2...max(root,2) where n % div == 0 {
    ...
}

However, at this point your solution with the regular for loop is more readable.

answered Jan 8, 2016 at 11:09

Sergey Kalinichenko

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2 Comments

Tiza Over a year ago

Thank you for your fast answer. I did expect that the for-in-loop would not be executed at all. Fixing the upper bounds by finding the maximum of upper and lower bounds is semantically different. I will follow your advice and keep the regular for loop.

Sulthan Over a year ago

@Tiza However, note that the regular C-style loop will get removed from Swift in Swift 3.0 so if you rewrite it now, it will save you time later. See github.com/apple/swift-evolution/blob/master/proposals/… You might want to use another for loop variant or a while loop.

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