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Xcode – How I need to use isMultiple(of: ) to print even indexed characters?

So my task is to only print even indexed characters using for-in loop.

Create a for-in loop that will loop through alphabet. Inside the loop, print every other letter by continuing to the next iteration if you are on a letter you do not wish to print. (Hint: You can use the isMultiple(of:) method on Int to only print even indexed characters).

Here’s my code:

let alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

for letter in alphabet {
    if letter == "F" { // letter I wish not to print
        continue
    }

    // How do I need to use isMultiple(of: ) method?

    print(letter)
}
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3

Answers


  1. Chosen as BEST ANSWER
    0

    I found out, that we need to put 'print' statement inside 'if' statement, because it will print even indexes (what task says).

    let alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

for (index, letter) in alphabet.enumerated() {
    if letter == "F" {
        continue
    }
    if index.isMultiple(of: 2) {
        print(letter)
    }
}

    But if we put 'continue' inside if statement, we will skip even indexes. So the odd ones will be printed. Am I right?


  2. 0

    Here’s how you can do it.

    let alphabets = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

for (index, letter) in alphabets.enumerated() {
    if index.isMultiple(of: 2) {
        continue
    }
    print(letter)
}
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  3. 0

    The swifty way is enumerated()– to get indexed characters – in conjunction with a where clause.

    The where clause skips the current iteration if the condition is false.

    let alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for (index, letter) in alphabet.enumerated() where index.isMultiple(of: 2) {
   print(letter)
}

    This prints "A C E …"

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