Correctly Parsing JSON in Swift 3

Question

I\'m trying to fetch a JSON response and store the results in a variable. I\'ve had versions of this code work in previous releases of Swift, until the GM version of Xcode 8

Answer 1

Swift has a powerful type inference. Lets get rid of "if let" or "guard let" boilerplate and force unwraps using functional approach:

1. Here is our JSON. We can use optional JSON or usual. I'm using optional in our example:

    let json: Dictionary? = ["current": ["temperature": 10]]

2. Helper functions. We need to write them only once and then reuse with any dictionary:

    /// Curry
    public func curry(_ f: @escaping (A, B) -> C) -> (A) -> (B) -> C {
        return { a in
            { f(a, $0) }
        }
    }

    /// Function that takes key and optional dictionary and returns optional value
    public func extract(_ key: Key, _ json: Dictionary?) -> Value? {
        return json.flatMap {
            cast($0[key])
        }
    }

    /// Function that takes key and return function that takes optional dictionary and returns optional value
    public func extract(_ key: Key) -> (Dictionary?) -> Value? {
        return curry(extract)(key)
    }

    /// Precedence group for our operator
    precedencegroup RightApplyPrecedence {
        associativity: right
        higherThan: AssignmentPrecedence
        lowerThan: TernaryPrecedence
    }

    /// Apply. g § f § a === g(f(a))
    infix operator § : RightApplyPrecedence
    public func §(_ f: (A) -> B, _ a: A) -> B {
        return f(a)
    }

    /// Wrapper around operator "as".
    public func cast(_ a: A) -> B? {
        return a as? B
    }

3. And here is our magic - extract the value:

    let temperature = (extract("temperature") § extract("current") § json) ?? NSNotFound

Just one line of code and no force unwraps or manual type casting. This code works in playground, so you can copy and check it. Here is an implementation on GitHub.