Try and catch around unwrapping line? Swift 2.0, XCode 7

Question

I have the following unwrapping line in my code:

UIApplication.sharedApplication().openURL((NSURL(string: url)!))

Sometimes there occurs this f

Answer 1

No, this is not what try and catch are for. ! means "if this is nil, then crash." If you don't mean that, then don't use ! (hint: you very seldom want to use !). Use if-let or guard-let:

if let url = NSURL(string: urlString) {
    UIApplication.sharedApplication().openURL(url)
}

If you already have a try block and want to turn this situation into a throw, that's what guard-let is ideal for:

guard let url = NSURL(string: urlString) else { throw ...your-error... }
// For the rest of this scope, you can use url normally
UIApplication.sharedApplication().openURL(url)