Binary operator '<' cannot be applied to two 'Int?' operands

Question

Good evening lovely community,
this is my first post, please have mercy, if I do something wrong.
I know there are some similar questions here, but I doesn\'t under

Answer 1

This error can also occur if you are comparing a custom type (ie: struct or class) for which you haven't implemented the Comparable protocol.

Answer 2

Well, by using guard statement you can check if both values are not nil, and converting it to Int typ

    guard let value_one = Int(goalPlayerOne), let value_two = Int(goalPlayerTwo) else {
        print("Some value is nil")
        return
    }

so you can safely compare two values.

    if value_one < value_two {
       //Do something
    }

Answer 3

If you look at the declaration for Int's initializer that takes a String, you can see by the ? after init that it returns an optional:

convenience init?(_ description: String)

This means you have to unwrap it before you can do most things with it (== is an exception, since the Optional type has an overload for that operator).

There are four main ways to unwrap your optionals:

1: If let

if let goalOne = Int(someString) {
    // do something with goalOne
}

2: Guard let

guard let goalOne = Int(someString) else {
    // either return or throw an error
}

// do something with goalOne

3: map and/or flatMap

let someValue = Int(someString).map { goalOne in
    // do something with goalOne and return a value
}

4: Provide a default value

let goalOne = Int(someString) ?? 0 // Or whatever the default value should be

If you unwrap all your optionals, you'll be able to compare them as you expect.