CODEWITHSUNDEEP
swiftfor-looprefactoringoperatorsstride
user3352185
user3352185

Reputation: 87

Swift 2.2 decrementing specific for loop in Swift 3

I have the task to refactor an iOS app to Swift 3. However, there is a for loop, in C-style, that does more than just looping an array backwards (it's mandatory to be backwards).

This is a sample code. The principle is the same.

let array = ["hello", "world", nil, "foo", nil, "bar", "Peter Griffin"]
var threeLetterWords = 0
for var i = array.count-1; i >= 0 && array[i].characters.count == 3; --i, ++threeLetterWords { }
print("Found words: \(threeLetterWords)") // should say `Found words: 2`

I tried with stride(from:through:by:) but I can't increment threeLetterWords as it seems important to increment it in the loop. Any ideas?

Upvotes: 0

Views: 433

Answers (4)

Leo Dabus
Leo Dabus

Reputation: 236478

You can use your array indices reversed and add a where clause for the characters count:

let array = ["hello", "world", nil, "foo", nil, "bar", "Peter Griffin"]
var threeLetterWords = 0

for index in array.indices.reversed() where array[index]?.characters.count == 3 {
    threeLetterWords += 1
}

print("Found words: \(threeLetterWords)") // should say `Found words: 2`

Upvotes: 1

vacawama
vacawama

Reputation: 154691

Your code isn't counting the number of 3-letter words in the array. It is counting the number of 3-letter words at the end of the array. It will return 0 for your sample input array.

When a C-style for loop is very complex, the final fallback solution is to translate it to a while loop. Any C-style for loop can be mechanically converted into an equivalent while loop, which means you can do it even if you don't fully understand what it is doing.

This for loop:

for initialization; condition; increment {
    // body
}

is equivalent to:

initialization
while condition {
    // body
    increment
}

So, your code is equivalent to:

let array = ["hello", "world", nil, "foo", nil, "bar", "Peter Griffin"]
var threeLetterWords = 0

var i = array.count - 1
while i >= 0 && array[i]?.characters.count == 3 {
    i -= 1
    threeLetterWords += 1
}
print("Found words: \(threeLetterWords)") // says `Found words: 0`

Here is how to use a for loop and guard to do the equivalent of your code:

let array = ["hello", "world", nil, "foo", nil, "bar", "Peter Griffin"]
var num3LetterWords = 0

for word in array.reversed() {
    guard word?.characters.count == 3 else { break }
    num3LetterWords += 1
}

print(num3LetterWords)

Upvotes: 1

Alexander
Alexander

Reputation: 63369

Everybody here is very unnecessarily complicating this.

let words = ["hello", "world", nil, "foo", nil, "bar", "Peter Griffin"]

var num3LetterWords = 0

for word in words.reversed() {
    if (word?.characters.count == 3) { num3LetterWords += 1 }
}

print(num3LetterWords)

Upvotes: 0

mqz.kim
mqz.kim

Reputation: 1039

//for var i = array.count-1; i >= 0 && array[i].characters.count == 3; --i, ++threeLetterWords { }

for i in stride(from: (array.count-1), through: 0, by: -1) {
    threeLetterWords += 1

    if (array[i]?.characters.count == 3) {
        break
    }
}

Upvotes: 2

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