Comments by hiwa doski (Top 47 by date)

i fixed it by changing
f1 = (FragmentA) manager.findFragmentById(R.id.main_fragment_container);
to
f1 = (FragmentA) fragment;

and it worked but i don't if this is the correct way of doing it.

thanks, i update it.

yeah i got it, thank you guys.

aha okay got it, thanks.

thank you for your will explained answer, but i still have some confusion about how/when inheriting is done i read the doc you sent me and it kept saying it inherits all field from super class so why the "e" reference variable `name` doesn't have value of Java as it inherits from it, or it does not inherit values ?

oh yeah that make sense, but how to set employee to the new person i already have as an Employee reference variable can't point to a person object ?

Okay thanks.

Yes that works thanks, but why .text doesnt work ?

i am watching a course and it explain things, but sometimes you won't just get a thing and have to ask!

i got the answer from other people which was a simple thing and i understood very quickly and i didn't had to study the whole Kotlin language to understand a simple thing.

no that didn't worked, it still says required !Editable found String

how about giving me a simple answer?

Thanks, that worked.

okay thanks.

yeah, I think there is because there is a number 3 line that show's in my IDE which is blank, kudos to you for figuring it out and thanks for your help, but one last question what i can do with that to make it not throw exception I already put '\n' in delimiter so what else can be done about it ?

my data is just like i mentioned it's a text file containing
P1,19
P2,25

your code worked and printed them but at the end i still get an exception !
i wrote your code like this

try (Scanner scanner = new Scanner((new FileReader("names-with re.txt")))) {
            scanner.useDelimiter(",|\\n");
            while (scanner.hasNextLine()) {
                String s = scanner.next();
                scanner.skip(scanner.delimiter());
                int age = scanner.nextInt();
                System.out.println(s);
                System.out.println(age);
            }
        }

and i still get NoSuchElementException with line `String s = scanner.next();', why is that ?

Thank you so much for your help.

i think it's okay but will take a look at it and thank you.

but didn't you already did it?

"roror" , because it has three duplicated "r" and "foo" only have two duplicated "o" so "roror" should be returned.

Thank you so much.

okay, will do that.

i don't think it matters, as long as there is no word with more than 4 duplicated characters.

it's 4 "s"

noh I'm not getting paid and it's not homework, it's just a problem I found online and want to know the solution for it, and i tried many solutions and i spent much time on it
my solutions were like this
1-

public static String LongestWord() {
        String[] words = {"Apple", "cheese", "Mississippi", "beer", "hand", 
        "sandwich", "kadabra"};

        String maxLength = words[0];
        String newString = "";


        for (String str : words) {

            for (int i = 0; i < str.length(); i++) {
                {
                  for (int j=i+1; j<str.length(); j++){
="" if="" (str.charat(i)="=" str.charat(j)){
="" newstring+="str.charAt(i);
" }
="" }

="" (newstring.length()=""> maxLength.length()) {
                maxLength = newString;
            }
        }

        return maxLength;
    } 

// the problem with the above solution is, it doesn't go through all characters of string it starts with "i" and there may be a duplicate before "i" and it won't detect it so it doesn't work.

2- Show   Expand   Copy Code

public static String LongestWord() {        String[] words = {"Apple", "cheese", "Mississippi", "beer", "hand", "sandwich", "kadabra"};        String maxLength = words[0];        String newString = "";        for (String str : words) {            for (int i = 0; i < str.length(); i++) {                {                  for (int j=str.length()-1; j>=0; j--){                      if (str.charAt(i) == str.charAt(j)){                          newString+=str.charAt(i);                      }                    }                }            }            if (newString.length() > maxLength.length()) {                maxLength = newString;            }        }        return maxLength;    }  

//this one is closer to work but it still has a problem and that is when for example "i=0" and "j" reaches 0 then it will be equal because it's the same letter but not a duplicate, and i tried many others too, but nothing worked.

That's a very good explanation, thanks.

and i thought if it got garbage collected it wasn't gonna get executed, so that's why i said it didn't got garbage collected.

Okay, thanks for your explanation.

because the method got executed!

i thought the garbage collector will immediately start and remove it, but it turned out that I was wrong.

if garbage collector is not related to this then i don't know what is ?

yeah i know, but i thought it will happen before that.

what you mean by "when your app exits"

Thanks.

Okay thank you.

explained very well, thanks.

Okay thank you.

is this the same when you declare a variable in a normal for loop like -> for (int i=0; i<10;i++)

okay thanks, but another question let's say child class contains method bar() if the reference variable of parent class point to the child object, and that child object contain the bar() method then why we can't call it with parent reference?