Binary operator '~=' cannot be applied to operands of type 'String' and 'String?'

Question

I have a simple switch-statement that is not that simple.

switch(bubble?.name){ //bubble is SKPhysicsBody
    case "largeBubble": // <= error
        newBubbleSize = "medium"
        break;
    default:
        newBubbleSize = "large"
        break;
}

Here I get error that I mentioned in title Binary operator '~=' cannot be applied to operands of type 'String' and 'String?'. And I have no clue why it is a problem that one of them is an optional.

jtbandes · Accepted Answer · 2015-06-30T18:15:26.983

8

Because of Optional Chaining, bubble?.name has type String?. You have a few options:

Use "largeBubble"? in your case expression (Swift 2+ only).
Check for nil before doing your switch, so the switch argument will be a String instead of String?.
Use bubble!.name (or bubble.name if it's a SKPhysicsBody!) if you are absolutely sure that it won't be nil

edited Jun 30 '15 at 18:15

answered May 30 '15 at 18:13

jtbandes

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Option #4: `switch(bubble?.name ?? "unknown") { ... }`. – Martin R May 30 '15 at 18:14
Sure, but then you will be doing an unnecessary comparison with the string "unknown" in some cases :) Of course, it's probably so fast that it doesn't matter... – jtbandes May 30 '15 at 18:15
`bubble!.name` didn't work - well, only in combination with `some()`. Thought that's why there's always a default as well ... if name is not set. – Jurik May 30 '15 at 18:22
@Jurik exactly what type is `bubble`? (`SKPhysicsBody` or `SKPhysicsBody?` or perhaps `!`) – jtbandes May 30 '15 at 18:23
It's `bodyA` or `bodyB` from `SKPhysicsContact`. – Jurik May 30 '15 at 18:26
`bodyA` and `bodyB` are both `SKPhysicsNode!`. I am not 100% sure, but I suspect these are really never nil, so it's safe to just use `bubble.name`. That is the solution I would recommend. – jtbandes May 30 '15 at 18:30

ABakerSmith · Answer 2 · 2015-05-31T09:25:09.603

3

As @jtbandes said, the problem is the result of bubble?.name having type String?. An alternative solution, to the ones given, is to override the ~= operator to take a String? and String as arguments, for example:

func ~= (lhs: String, rhs: String?) -> Bool {
    return lhs == rhs
}

Or, to make it more generic:

func ~= <T: Equatable>(lhs: T, rhs: T?) -> Bool {
    return lhs == rhs
}

edited May 31 '15 at 09:25

answered May 30 '15 at 18:19

ABakerSmith

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`Equatable` would be sufficient :) – Martin R May 30 '15 at 18:21

score 1 · Answer 3 · answered Sep 19 '15 at 08:45

“Values are never implicitly converted to another type. If you need to convert a value to a different type, explicitly make an instance of the desired type.”

“let label = "The width is "
let width = 94
let widthLabel = label + String(width)”

Here width is an Integer type it has been converted to String by String(Int) function

Ranjithkumar · Answer 4 · 2015-09-24T05:14:21.560

In Swift 2.0 (sender: UIButton) then ! for the switch string works.

let operation = sender.currentTitle!

In Swift 2.0, (sender: AnyObject) then !! for the switch string works.

let operation = sender.currentTitle!!

the mistake which i made is using AnyObject instead of UIButton.

@IBAction func operate(sender: AnyObject) {
        let operation = sender.currentTitle!!

        if userIsInTheMiddleOfTypeingNumber{
            enter()
        }

        switch operation{
            case "×":
                if operandStack.count >= 2{
                    displayValue = operandStack.removeLast() * operandStack.removeLast()
                }

                break
//            case "÷":
//                break
//            case "+":
//                break
//            case "−":
//                break
           default:
                break
        }

    }

Binary operator '~=' cannot be applied to operands of type 'String' and 'String?'

4 Answers4