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iOS Swift Functions and Optionals Optionals Optional Chaining

Posted by Stanley Lim
Stanley Lim
209 Points

Why is 2:51 unsafe with the bang unwrapping?

Amit mentioned that bang unwrapping is unsafe, and I could understand why cause we can never expect valuables to have a value, however, in 2:51, culprit has already did an optional checking in line 21. So if culprit.toInt()! in line 22 can be reached, it definitely has a value already, therefore being safe? No?

Would appreciate if someone can clear this up :)

Thanks in advance

2 Answers

mustangxy
mustangxy
3,212 Points
mustangxy
mustangxy
3,212 Points

In the case you described yes, it is safe to use force-unwrapping. It is not in all those situations where you don't first make sure (with optional binding, for example) that the value exists. If you wrote:

let culprit = findApt("102")
culprit.toInt()!

then you'd be writing unsafe code, because you didn't perform any sort of check before force-unwrapping.

Eduardo Calvachi
Eduardo Calvachi
6,049 Points
Eduardo Calvachi
Eduardo Calvachi
6,049 Points

For Swift 2.1, since the optional was safely unwrapped and the value inside the conditional statement can be casted as an int:

if let culprit = findApt("104") {
        sendNoticeTo(apt: Int(culprit)!)
    }