CODEWITHSUNDEEP
stringswiftintoption-type
sdnts
sdnts

Reputation: 728

Additional characters when unwrapping optional in Swift

I'm extremely new to Swift, and I'm trying to build a simple calculator to test the waters. I have a UILabel with name 'display', and I want to store its value as an int in a variable called 'numA'. However, when I did that, the variable numA gets the value nil.

I discovered why this happened, because when I simply print the display text in the console:

println("\(display.text!)")

I get this: " 6" instead of "6". Notice the additional space appended at the beginning. This is why I cannot do this:

var numA = display.text!.toInt()

What am I doing wrong?

Thanks

Upvotes: 0

Views: 183

Answers (1)

vacawama
vacawama

Reputation: 154741

The leading space might be there because you are initializing your display to " " (a single space) instead of to "" (empty string). Even if you have spaces in your display.text, you can trim them out before calling toInt().

Note that since toInt() returns an optional, you should use optional binding to unwrap it to prevent app crashes. I recommend:

if let numA = display.text?.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet()).toInt() {
    // use numA
}

Note the use of ? after text instead of !. This will keep your app from crashing if display.text is nil. This is an example of an optional chain; if display.text is nil, the entire chain will be nil.

This can be further simplified by leaving out NSCharacterSet which Swift can infer from the call:

if let numA = display.text?.stringByTrimmingCharactersInSet(.whitespaceCharacterSet()).toInt() {
    // use numA
}

Upvotes: 1

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