Swift-fatal error: Array index out of range

Question

How can I fix this code so I no longer get an error. The error I get is fatal error: Array index out of range. The code checks how many "x" and "o" there are in a string

var input = "xxxoxoooxoxo"

var inputString = Array(input.characters)

var XString = ""

var OString = ""

for var i = 0; i <= inputString.count; ++i {

    if inputString[i] == "x" {

        XString.append(inputString[i])

    }

    else if inputString[i] == "o" {

        OString.append(inputString[i])

    }

}

if XString.characters.count == OString.characters.count {

    print("equal")

}

else {

    print("Not Equal")

}

Thanks for the help.

Keep in mind the standard for loop will be deprecated in swift 2.2 and removed in 3.0 — ronatory
– ronatory, Commented Mar 20, 2016 at 7:11

dasdom · Accepted Answer · 2016-03-20 06:55:00Z

3

Replace this

for var i = 0; i <= inputString.count; ++i

with this:

for var i = 0; i < inputString.count; ++i

Arrays are zero indexed. That means the first element has the index 0. The second element has index 1. ... . The last element has the index array.count-1.

answered Mar 20, 2016 at 6:55

dasdom

14.1k2 gold badges50 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

nielsbot · Accepted Answer · 2016-03-20 07:35:13Z

3

@dasdom is correct. but here is a more Swift-y way to do it:

// input string
var input = "ooooxxxxxxxoxoooxoxo" 

// difference between number of x's and number of o's
var diff = 0 

// perform this function for each character in `input.characters`:
input.characters.forEach {
    switch $0 // $0 is the argument to our function (unnamed)
    {
        case "x": diff += 1          // x's are +1
        case "o": diff -= 1          // o's are -1
        default: fatalError()        // we expect only 'x' or 'o'... if we get something else, just crash
    }
}

// use the ternary (?:) operator to print the answer.
// (<result> = <condition> ? <if-true value> : <if-false value>)
print( diff == 0 ? "equal" : "not equal" )

A shorter version, inspired by @Eendje's answer :)

print( input.characters.reduce(0) { $0 + [ "x":1, "o":-1 ][ $1 ]! } != 0 ? "not" : "", "equal" )

edited Mar 20, 2016 at 7:35

answered Mar 20, 2016 at 6:59

nielsbot

16k4 gold badges53 silver badges75 bronze badges

6 Comments

Eendje Over a year ago

Haha, nice one! I really like the use of a dictionary here. I would use ?? 0 instead of implicitly unwrapping it though :) (Maybe I should add that in my answer! :P)

dasdom Over a year ago

The second version is quite interesting, but I think it's to much for someone struggling with the number of elements in an array. ;)

nielsbot Over a year ago

I used force unwrap to catch "invalid" characters... but if you don't care about that, then ?? 0 works, sure.

Eendje Over a year ago

I wonder, lets say input has over 1000 characters. How expensive is it to create a dictionary for every iteration? Do you happen to have any idea?

nielsbot Over a year ago

hopefully, since it's a literal it won't be created every time? but if that's a concern, the dictionary could be created outside the closure...

|

Eendje · Accepted Answer · 2016-03-20 07:22:13Z

2

While everyone is giving you an alternative, I thought this would be a nice addition as well:

var input = "xxxxxoxoooxoxo"

You can get the count by using filter:

let xCount = input.characters.filter { $0 == "x" }.count
let oCount = input.characters.filter { $0 == "o" }.count

print( xCount == oCount ? "equal" : "not equal" )

In your case, using reduce would be more efficient and it's shorter:

let result = input.characters.reduce(0) { $0 + ($1 == "x" ? 1 : -1) }

print(result == 0 ? "equal" : "not equal") // not equal

This is only if you're sure input only contains x and o. If not, then you'll have to change the evaluation to:

{ $0 + ($1 == "x" ? 1 : $1 == "o" ? -1 : 0) }

edited Mar 20, 2016 at 7:22

answered Mar 20, 2016 at 7:01

Eendje

8,8931 gold badge31 silver badges31 bronze badges

Comments

John Difool · Accepted Answer · 2016-03-20 06:59:48Z

0

You are enumerating pass the range of the array. Use this instead.

for var i = 0; i < inputString.count; ++i {

Or you can use this more idiomatic code:

var input = "xxxoxoooxoxo"


var xString = 0

var oString = 0

for c in input.characters {

    if c == "x" {

        xString++

    }

    else if c == "o" {

        oString++

    }

}

if oString == xString {

    print("equal")

}

else {

    print("Not Equal")

}

answered Mar 20, 2016 at 6:59

John Difool

5,7627 gold badges49 silver badges86 bronze badges

Collectives™ on Stack Overflow

Swift-fatal error: Array index out of range

4 Answers 4

Comments

6 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

6 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related