How to solve “String interpolation produces a debug description for an optional value; did you mean to make this explicit?” in Xcode 8.3 beta?

Question

Since beta 8.3, zillions warnings \"String interpolation produces a debug description for an optional value; did you mean to make this explicit?\" appeared in my code.

Answer 1

See Ole Begeman's fix for this. I love it. It creates a ??? operator which you can then use like this:

var someValue: Int? = 5
print("The value is \(someValue ??? "unknown")")
// → "The value is 5"
someValue = nil
print("The value is \(someValue ??? "unknown")")
// → "The value is unknown"

Answer 2

Two easier ways of dealing with this issue.

Option 1:

The first would be by "force-unwrapping" the value you would like to return using a bang (!)

var someValue: Int? = 5
print(someValue!)

Output:

5

Option 2:

The other way, which could be the better way - is to "safely-unwrap" the value you want returned.

var someValue: Int? = 5

if let newValue = someValue {
    print(newValue)
}

Output:

5

Would recommend to go with option 2.

Tip: Avoid force unwrapping (!) where possible as we are not sure if we will always have the value to be unwrapped.

Answer 3

Create an interpolation method that accepts an optional generic Type with an unnamed parameter. All your annoying warnings will magically disappear.

extension DefaultStringInterpolation {
  mutating func appendInterpolation<T>(_ optional: T?) {
    appendInterpolation(String(describing: optional))
  }
}

Answer 4

seems using String(describing:optional) is simplest.

default value ?? makes no sense for non-Strings e.g Int.
If Int is nil then you want the log to show 'nil' not default to another Int e.g. 0.

Some playground code to test:

var optionalString : String? = nil
var optionalInt : Int? = nil

var description_ = ""
description_ = description_ + "optionalString: \(String(describing: optionalString))\r"
description_ = description_ + "   optionalInt: \(String(describing: optionalInt))\r"

print(description_)

Output

optionalString: nil
optionalInt: nil

Answer 5

Double click on the yellow triangle displayed on line containing this warning. This will show FixIt with two solutions.

1. Use String(describing:) to silence this warning :

   Using this it will become String(describing:<Variable>)

   Eg. : String(describing: employeeName)

2. Provide a default value to avoid this warning :

   Using this it will become (<Variable> ?? default value)

   Eg.: employeeName ?? “Anonymous” as! String

Answer 6

Swift 5

My solution is making an extension which unwrap Optional object to Any.

When you log the object or print it out, you can see the actual object or <nil>⭕️ (combination from text and visual character). It's useful to look at, especially in the console log.

extension Optional {
    var logable: Any {
        switch self {
        case .none:
            return "<nil>|⭕️"
        case let .some(value):
            return value
        }
    }
}

// sample
var x: Int?
print("Logging optional without warning: \(x.logable)")
// → Logging optional without warning: <nil>|⭕️