How do you call a method that has an "if" statement within it from a class?

Question

class Countries {
    let name: String
    let size: Int
    let disasterRisk: Int
    let capital: String
    let continent: String

    init(name: String, size: Int, disasterRisk: Int, capital: String, continent: String) {
        self.name = name
        self.size = size
        self.disasterRisk = size
        self.capital = capital
        self.continent = continent
}
    func raiseAlarm(if atRisk: Int) {
        let scale = 30
        if scale > 30 {
            print("alert")
        }else if scale < 30 {
            print("low risk")
        }
    }

}

let countryStat = Countries(name: "NY", size: 4000, disasterRisk: 30, capital: "NYC", continent: "USA")
countryStat.raiseAlarm(if: 33)
print(countryStat.raiseAlarm(if: 44))

I am stuck!! please help

Answer 1 · 2018-02-02T11:22:24Z

February 2, 2018 11:22am

Hi there,

You are calling your method correctly with countryStat.raiseAlarm(if: 33) but the value you pass in is ignored.

Your method receives an integer with an external name of if and a local name of atRisk. You then do all your comparisons using a variable that is set internally to 30, scale. The method prints either "alert" or "low risk" depending whether scale is either above 30, or below 30. It is neither - scale is set to exactly 30 so the method will output nothing. The scenario of scale being equal to 30 isn't handled by your method. If you change else if to just else (with no additional condition) then the method will always output "low risk".

What you want to do is manage the output based on the input integer. Again, we'll handle the condition as being binary; the integer is either greater than 30 or it isn't. No other conditions are required; you may want to tweak the condition depending how you want the scenario of equal to 30 handled; is that an alert or a low risk?

Something like:

    func raiseAlarm(if atRisk: Int) {
        if atRisk > 30 {
            print("alert")
        }else {
            print("low risk")
        }
    }

I hope that helps,

Steve.

Answer 2 · 2018-02-02T17:47:55Z

February 2, 2018 5:47pm

Hi Steve,

many thanks for your help!

It works! thank your thorough explanation and helping me understand it a bit more

:)

Kavita

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Kavita Bargota

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How do you call a method that has an "if" statement within it from a class?

2 Answers

Steve Hunter

Steve Hunter

Kavita Bargota

Kavita Bargota

Steve Hunter

Steve Hunter

Kavita Bargota

Kavita Bargota