ist der fehler in meiner herangehensweise? warum klappt das nicht?

Frage zur Aufgabe

Gelöst

Erstelle eine Liste von Ganzzahlen.
Gin 20 Ganzzahlen über die Tastatur ein.
Erstelle eine Methode, um Zahlen sicher aus der Liste abzurufen:
int elementSicherAbrufen(ArrayList<Integer> liste, int index, int standardwert)
Die Methode muss ein Listenelement basierend auf seinem Index zurückgeben.
Wenn beim Abrufen eines Elements eine Ausnahme auftritt, muss diese abgefangen werden, und die Methode muss standardwert zurückgeben.

Anforderungen:

Das Programm muss 20 Zahlen von der Tastatur lesen.
Das Programm muss Daten auf dem Bildschirm anzeigen.
Die Methode elementSicherAbrufen muss ein Listenelement basierend auf seinem Index zurückgeben, es sei denn, es treten Ausnahmen in der Methode auf.
Die Methode elementSicherAbrufen muss standardwert zurückgeben, wenn Ausnahmen in der Methode auftreten. Fange Ausnahmen ab.
Die Methode elementSicherAbrufen darf keine Ausnahmen auslösen.

package de.codegym.task.task10.task1017; import java.io.BufferedReader; import java.io.*; import java.util.*; /* Eine Liste sicher abrufen */ public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); ArrayList<Integer> liste = new ArrayList<Integer>(); for (int i = 0; i < 20; i++) { int x = Integer.parseInt(reader.readLine()); liste.add(x); } System.out.println(elementSicherAbrufen(liste, 5, 1)); System.out.println(elementSicherAbrufen(liste, 20, 7)); System.out.println(elementSicherAbrufen(liste, -5, 9)); } public static int elementSicherAbrufen(ArrayList<Integer> liste, int index, int standardwert) { //schreib hier deinen Code int back = 0; Iterator<Integer> it = liste.iterator(); while(it.hasNext()){ if(it.next() == index){ back = it.next(); continue; } else if (!it.hasNext()){ back = standardwert; } } return back; } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

29 Mai 2020, 18:14

Take for an example an ArrayList with just these 5 values: {10, 20, 30, 40, 50} The way that your code works is that it searches each index of the passed in ArrayList for the number 'index' from the argument, and if it does not contain this number return the default number. So if the passed in argument was 3 with a default value of 35 the code would do this: does the number at index 0 (10) = 3? no does the number at index 1 (20) = 3? no does the number at index 2 (30) = 3? no does the number at index 3 (40) = 3? no does the number at index 4 (50) = 3? no so then it would return 35 The way that the code should work is it should get the number AT the specified index of the ArrayList and if an exception is thrown then return the default number. So, once again, if the passed in argument was 3 with a default value of 35 the code should do this:: the number at index 3 is 40. Return 40. if the index passed in was 5 instead of 3 then the return value would be 35 because accessing index 5 would throw an IndexOutOfRangeException and the default value would be returned instead.

hidden #10625598

Level 23

29 Mai 2020, 18:39

yea right, sorry i had a chaos with my tabs, i meant to upload this one.. public static int elementSicherAbrufen(ArrayList<Integer> liste, int index, int standardwert) { //schreib hier deinen Code int back = 0; for(int i = 0; i < liste.size(); i++){ if(liste.get(i) == index){ back = liste.get(i); }else{ back = standardwert; } } return back; } my output is 1, 7, 9 -> the default values.. but why

Guadalupe Gagnon

Level 37 , Tampa, United States

29 Mai 2020, 19:03

elementSicherAbrufen(ArrayList<Integer> liste, int index, int standardwert) ↑ index is the index number of the list. So if the list is {10, 20, 30, 40, 50} and index = 3 {10, 20, 30, 40, 50} ↑ 40 is in index 3

hidden #10625598

Level 23

29 Mai 2020, 19:11

exactly :D so where is my mistake, liste.get(i) is 3 at some point and then it will match with this index 3.. but yea so im giving it the wrong value right after.

Guadalupe Gagnon

Level 37 , Tampa, United States

29 Mai 2020, 19:21

i is the current loop index that is accessing each index of the list (which is very common inside for loops). This task however does not want you to do that. In my example, when i equals 3 list.get(i) would equal 40. The code you shared would compare: if(40 == 3) back = 40; This would not produce the correct results because 40 does not equal 3.

hidden #10625598

Level 23

29 Mai 2020, 19:25

i see, right, is it possible that we never did a excercise with this one.. so i just found on google indexOf() and it doesnt work aswell - how do i get this index number?

Guadalupe Gagnon

Level 37 , Tampa, United States

29 Mai 2020, 19:26

in the method it is provided as an argument. The code should use that argument. Simply put:

list.get(index)

hidden #10625598

Level 23

29 Mai 2020, 19:30

naah doesnt work tho if(liste.get(index) == index){ back = liste.get(i); }else{ back = standardwert; }

Guadalupe Gagnon

Level 37 , Tampa, United States

29 Mai 2020, 19:37

you are still comparing the wrong numbers. If the number at index 3 was 40: if(liste.get(index) == index) if(40 == 3 ) << this is what that would translate to This is incorrect and wont pass.