FAQ: Closures - Built-in Higher-Order Functions

Frequently Asked Questions Uncategorized
1

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This community-built FAQ covers the “Built-in Higher-Order Functions” exercise from the lesson “Closures”.

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This exercise can be found in the following Codecademy content:

Learn Intermediate Swift

FAQs on the exercise Built-in Higher-Order Functions

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2

The solution is:
let aNames = names.filter { $0.starts(with: [“A”]) }

print(aNames)

and long version:
let aNames = names.filter {(name) → Bool in

return names.starts(with: [“A”])

}

3

In step 1, the code:

let aNames = names.filter { $0.starts(with: [“A”]) }
print(aNames)

produces the exact same result as using:

let aNames = names.filter { $0.starts(with: “A”) }
print(aNames)

yet only the former is accepted.

  1. What is the argument: [“A”]? I know that obviously each string is being represented as an array of characters in order to test if the first character is “A”, but why are the brackets necessary? The data type of the first char in the string, or perhaps more precisely the first element in the array as the string is being tested, would still be char and not [Char] so why the brackets?

  2. Is there any meaningful differences between the two syntaxes above, or is this just Codecademy being finicky when it comes to checking the code? As mentioned both work identically, but I don’t know if there is any best practice differences I’m missing.

Thanks in advance for the clarification!