JuniorTheoryCommon

What does the program print?

SWIFT

enum ParseError: Error {
    case badFormat
}
func parse(_ text: String) throws -> Int {
    guard let n = Int(text) else {
        throw ParseError.badFormat
    }
    return n
}
let a = (try? parse("8")) ?? -1
let b = (try? parse("8x")) ?? -1
print(a, b)

8 -1

Expecting 8 0: thinking try? turns an error into zero. It gives nil, and only ?? -1 decides what stands in its place.
Expecting a crash: confusing try? with try!. Only try! stops the program on an error; try? turns the error into nil.

What would b hold if the parentheses around try? parse("8x") were removed?
When is try? the wrong choice because the caller needs the reason for the error?

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JuniorTheoryCommon

What does the program print?

SWIFT

enum ParseError: Error {
    case badFormat
}
func parse(_ text: String) throws -> Int {
    guard let n = Int(text) else {
        throw ParseError.badFormat
    }
    return n
}
let a = (try? parse("8")) ?? -1
let b = (try? parse("8x")) ?? -1
print(a, b)

8 -1

Expecting 8 0: thinking try? turns an error into zero. It gives nil, and only ?? -1 decides what stands in its place.
Expecting a crash: confusing try? with try!. Only try! stops the program on an error; try? turns the error into nil.

What would b hold if the parentheses around try? parse("8x") were removed?
When is try? the wrong choice because the caller needs the reason for the error?

Back to questions Open topic in roadmap →Answer it yourself — Quiz