Optional value in swift 3 Optimised way - swift

Question

I have create class Testing model which has 4 dataMember it should not be null when accessing (means return default value)
extension Double {
/// Rounds the double to decimal places value
func roundTo(places:Int = 2) -> Double
{
let divisor = pow(10.00, Double(places))
return (self * divisor).rounded() / divisor
}
}
class TestingModel{
var id : String!
var name : String! = "abc" /*It is not working*/
var price : Double! = 0.00
var uniqueId : Int! = 1
/**
* Instantiate the instance using the passed dictionary values to set the properties values
*/
init(dictionary: [String:Any])
{
id = (dictionary["id"] as? String) ?? "" //I dont want to do like this way
name = dictionary["name"] as? String
price = (dictionary["price"] as? Double)?.roundTo() ?? 0.00
uniqueId = dictionary["unique_id"] as? Int
}
}
let t:TestingModel = TestingModel.init(dictionary: ["x id" : "x012y12345z45","x name":"test1","x price":100.0,"uniqueId":1236.0])
let testString = "Jd " + t.id
print(testString) //Perfect
print(t.name)
print(t.price) /* Only one decemal point is printed */
Getting Output
Jd
nil
0.0
Expected output
Jd
abc /Should return abc instead of nil/
0.00 /Two decimal point complulsury/
What i actually mean in
if i assign nil value to variable then it should remain with its default value without writing this Optional chaining ?? "abc" in constructor

Answer 1

price is a Double type and what you are asking to do is to print that double value to 2 decimal places. Then you should make use of the following.
let a = 0.0
print(String(format: "%.2f", a))
this prints:
0.00
If you are planning to round it to decimal places, then also the above code will return that. But if you need it to round and return a double type then you can check this answer
Based on your updated question, I suggest to use the model as follows:
class TestingModel{
var id : String = ""
var name : String = "abc"
var price : Double = 0.0
var uniqueId : Int = 1
/**
* Instantiate the instance using the passed dictionary values to set the properties values
*/
init(dictionary: [String:Any])
{
id = (dictionary["id"] as? String) ?? ""
name = dictionary["name"] as? String ?? "abc"
price = (dictionary["price"] as? Double) ?? 0.0
uniqueId = dictionary["unique_id"] as? Int ?? 1
}
}

Answer 2

You seem to have asked two different questions here. The first one regarding doubles have already been answered by adev. I will answer the second one, which is:
if i assign nil value to variable then it should remain with its default value without writing this Optional chaining ?? "abc" in constructor
If you want to do this then it means that the variable shouldn't be optional at all, as nil isn't one of its valid values. Make the variable a non-optional type and give it a default value.
class TestingModel{
var id : String = ""
var name : String = "abc"
var price : Double = 0.00
var uniqueId : Int = 1
}
You can't really avoid using ?? in the constructor because of the nature of dictionaries. They will always return a nil value if the key does not exist. You have to check it. It does not make sense even if this is possible anyway. Imagine something like this:
someVariable = nil // someVariable is now 0
This is extremely confusing. someVariable is 0, even though it appears that nil is assigned to it.
A workaround will be to add a dictionary extension. Something like this:
extension Dictionary {
func value(forKey key: Key, defaultValue: Value) -> Value {
return self[key] ?? defaultValue
}
}
But I still recommend that you use ?? instead.