Why wont this conditional work in Swift 3?

Question

Hey prior to Swift 3 I previously used :

if CustomUser.sharedInstance.numberOfCredits > 0 {
}

Where numberOfCredits is Double?. Now I get the error: Binary operator '>' cannot be applied to operands of type 'Double?' and 'Int'. Why is this? So I change 0 to 0.0 and get Binary operator '>' cannot be applied to operands of type 'Double?' and 'Double'. Why doesn't this line work in Swift 3?

Can you display your shared instance and function? May be there is problem of optionals. — Ashish Kakkad, Sep 24 '16 at 10:38
try this: `if CustomUser.sharedInstance.numberOfCredits! > 0.0` — Bista, Sep 24 '16 at 10:43
Please check [this](http://stackoverflow.com/a/39427736/6541007). — OOPer, Sep 24 '16 at 10:43

score 5 · Answer 1 · answered Sep 24 '16 at 10:44

This doesn't work because your numberOfCredits is an Optional (Double?). Should be: 0121 Remove Optional Comparison Operators.

Do it like this:

if let v = CustomUser.sharedInstance.numberOfCredits, v > 0 {
}

P.S.: Apart from that: numberOfCredits sounds like it should be an Int, not a Double? And maybe it shouldn't be an Optional either - if there are no credits, it should be 0? Hard to say w/o knowing what you are doing.

score 2 · Answer 2 · answered Sep 24 '16 at 10:49

Swift 3 has removed comparisons between an optional and a non-optional with the same base type. You have three options to handle this:

Explicitly unwrap the value with ! if you know that it is non-nil,
Use if let construct to unwrap the optional,
Use nil coalescing operator

Here is how you can use the third option:

if CustomUser.sharedInstance.numberOfCredits ?? 0 > 0 {
}

Note: There is a possibility that numberOfCredits is optional by omission - for example, because sharedInstance is defined as optional without automatic unwrapping. In this case it is better to fix the underlying problem, rather than using ?? or ! as a workaround.

Wilson · Answer 3 · 2016-09-24T13:23:26.533

struct CustomerUser {
  static var sharedInstance = Instance()
}

struct Instance {
  var numberOfCredits: Double?
}

let credits = CustomerUser.sharedInstance.numberOfCredits ?? 0.0

if credits > 0 {
  print("credits greater than zero")
}

numberOfCredits is an Optional Double (Double?) means it may contain a nil or a Double value.

That's why compiler wouldn't know what to do if numberOfCredits were nil.

if CustomUser.sharedInstance.numberOfCredits > 0 // wtf

In order to convert Double? to Double, you might use the ?? operator which not only will convert in case numberOfCredits contains a Double value but in case numberOfCredits contains nil will return 0.0, how cool that is?

let credits = CustomerUser.sharedInstance.numberOfCredits ?? 0.0

Why wont this conditional work in Swift 3?

3 Answers3