switch and cannot find variable in scope

Question

Why should I create a variable outside of switch/case? For example this code will have an error Cannot find 'size' in scope:

func sizeCheckNoVar(value: Int) -> String {
        switch value {
        case 0...2:
            let size = "small"
        case 3...5:
            let size = "medium"
        case 6...10:
            let size = "large"
        default:
            let size = "huge"
            
        }
        return size
    }

There is a default condition and AFIK all options are covered.

In the same time this code will be fine:

func sizeCheckVar(value: Int) -> String {
    var size: String
    switch value {
    case 0...2:
        size = "small"
    case 3...5:
        size = "medium"
    case 6...10:
        size = "large"
    default:
        size = "huge"
        
    }
    return size
}

PS I saw this question Cannot find variable in scope , but I want to know why instead of how to avoid

Because when you create a variable inside the case then the scope of the variable is within the case and you can’t access the variables outside the case — Raja Kishan
– Raja Kishan, Commented Nov 27, 2021 at 8:51
For the easy way returns the value direct from the case like case 0..2: return “small” … no need any variables — Raja Kishan
– Raja Kishan, Commented Nov 27, 2021 at 8:53

vadian · Accepted Answer · 2021-11-27 09:22:59Z

3

A pair of braces is called a scope.

In Swift (unlike some other languages) there is a simple but iron rule:

A variable declared inside a scope – in your particular case inside the switch statement – is visible in its own scope and on a lower level – like in your second example
It's not visible on a higher level outside the scope – like in your first example.

You can even declare size as constant because it's guaranteed to be initialized.

func sizeCheckVar(value: Int) -> String {
    let size: String
    switch value {
      case 0...2: size = "small"
      case 3...5: size = "medium"
      case 6...10: size = "large"
      default: size = "huge"
    }
    return size
}

However actually you don't need the local variable at all. Just return the values immediately

func sizeCheckVar(value: Int) -> String {
    switch value {
      case 0...2: return "small"
      case 3...5: return "medium"
      case 6...10: return "large"
      default: return "huge"
    }
}

Side note: The colon in a switch statement is also a kind of scope separator otherwise you would get errors about redeclaration of variables in the first example.

answered Nov 27, 2021 at 9:22

vadian

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2 Comments

submariner Over a year ago

How can I put the return value into a variable?

vadian Over a year ago

@submariner let myVariable = sizeCheckVar(value: 4)

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