An elegant way to ignore any errors thrown by a method

Question

I'm removing a temporary file with NSFileManager like this:

let fm = NSFileManager() fm.removeItemAtURL(fileURL) // error: "Call can throw but it is not marked with 'try'" 

I don't really care if the file is there when the call is made and if it's already removed, so be it.

What would be an elegant way to tell Swift to ignore any thrown errors, without using do/catch?

Answer 1

If you don't care about success or not then you can call

let fm = NSFileManager.defaultManager() _ = try? fm.removeItemAtURL(fileURL) 

From "Error Handling" in the Swift documentation:

You use try? to handle an error by converting it to an optional value. If an error is thrown while evaluating the try? expression, the value of the expression is nil.

The removeItemAtURL() returns "nothing" (aka Void), therefore the return value of the try? expression is Optional<Void>. Assigning this return value to _ avoids a "result of 'try?' is unused" warning.

If you are only interested in the outcome of the call but not in the particular error which was thrown then you can test the return value of try? against nil:

if (try? fm.removeItemAtURL(fileURL)) == nil {     print("failed") } 

---

Update: As of Swift 3 (Xcode 8), you don't need the dummy assignment, at least not in this particular case:

let fileURL = URL(fileURLWithPath: "/path/to/file") let fm = FileManager.default try? fm.removeItem(at: fileURL) 

compiles without warnings.