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iOS Swift Functions and Optionals Optionals Optional Chaining

Linh Nguyen
PLUS
Linh Nguyen
Courses Plus Student 984 Points
Posted by Linh Nguyen
Linh Nguyen
Courses Plus Student 984 Points

.toInt() is longer available and it's driving me insane

In the newest version of Swift, we no longer have the .toInt() method. The only way is to initialize Integer using constructor:

if let culprit = Int(findApt("102")?)

Then I get the error message:

"value of optional type 'String?' not unwrapped; did you mean to use '!' or '?'?"

So it suggested that I change the code to:

if let culprit = Int((findApt("102")?)!)

Even then, there's another error message that reads:

"optional chain has no effect, expression already produces 'String?'"

I can delete the Optional(?) and it will be okay as long as findApt() doesn't return a nil value.

What can I do to go around this?

Unsubscribed User
Unsubscribed User
Courses Plus Student 22,129 Points

Linh,

Im Having the same problem, if I figure it out or find a solution link i'll post it as a reference.

1 Answer

Unsubscribed User
Unsubscribed User
Courses Plus Student 22,129 Points

So here is the solution that managed to work for me. Still trying to figure out a way to minimize the "if-let" statements. Hope this helps for the time being

if let culprit = findApt("101"){
    if let aptNumber = Int(culprit){
        sendNoticeTo(aptNumber)
    }
}