Given a list of Strings, return an Array containing the same Strings in the same order

Question

I can't figure out why the last value of stringList (ie, Banana) is not being printed out or stored in arrayForTheList for that matter.

Given a list of Strings, return an array containing the same Strings in the same order

list2Array( ["Apple", "Orange", "Banana"] )         ->  {"Apple", "Orange", "Banana"}
list2Array( ["Red", "Orange", "Yellow"] )           ->  {"Red", "Orange", "Yellow"}
list2Array( ["Left", "Right", "Forward", "Back"] )  ->  {"Left", "Right", "Forward", "Back"}

public String[] list2Array(List<String> stringList) {

    String[] arrayForTheList = new String[(stringList.size())];

    for (int i = 0; i < stringList.size() ; i++) {
        arrayForTheList[i] = stringList.remove(0);
        System.out.println(arrayForTheList[i]); 
    }

    return arrayForTheList;
}

You're incrementing i and you're deleting from the start of the list, so the elements are being moved towards the start of the list. — khelwood
– khelwood, Commented Jan 23, 2019 at 9:40
Why don't use String[] arrayForTheList = (String[]) stringList.toArray();? — Dang Nguyen
– Dang Nguyen, Commented Jan 23, 2019 at 9:41
Why don't you try something like: String[] array = new String[stringList.size()]; return stringList.toArray(array); — Stultuske
– Stultuske, Commented Jan 23, 2019 at 9:41
You can use remove, becuse it decrements the size of the list that you use dynamically in the loop — azro
– azro, Commented Jan 23, 2019 at 9:44
I originally marked as a duplicate of stackoverflow.com/questions/7969023/from-arraylist-to-array. Not quite the same, but the linked SO question offers a better conversion mechanism — Brian Agnew
– Brian Agnew, Commented Jan 23, 2019 at 9:45

Eran · Accepted Answer · 2019-01-23 09:41:54Z

1

You shouldn't remove elements from the List inside your loop. You can clear the List after the loop if you have to.

for (int i = 0; i < stringList.size() ; i++) {
    arrayForTheList[i] = stringList.get(i);
    System.out.println(arrayForTheList[i]); 
}
stringList.clear();

answered Jan 23, 2019 at 9:41

Eran

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talex · Accepted Answer · 2019-01-23 09:46:31Z

1

Replace

for (int i = 0; i < stringList.size() ; i++)

to

for (int i = 0; i < arrayForTheList.length ; i++)

Keep in mind that your code clear List which it receive as parameter.

answered Jan 23, 2019 at 9:46

talex

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1 Comment

Jeppe Over a year ago

Yeah, the problem is stringList.size() changes every iteration, because of stringList.remove(0).

Ben · Accepted Answer · 2019-01-23 09:57:49Z

1

You can simply implement this by using the toArray() method of lists. For example:

String[] arrayForTheList = (String[])stringList.toArray();

Hope this helps.

edited Jan 23, 2019 at 9:57

answered Jan 23, 2019 at 9:45

Ben

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2 Comments

Dang Nguyen Over a year ago

uhmm, you may forget to cast to String[] Type mismatch: cannot convert from Object[] to String[], or a better way is String[] arrayForTheList = stringList.toArray(new String[0]);

Ben Over a year ago

Yes you are correct, I forgot to cast to String[]. I updated my answer. Thanks!

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