ArrayList of int array in java

Question

I m new to the concept of arraylist  I ve made a short program that is as follows   ArrayList lt int   gt  arl new ArrayList lt int   gt     int a1    1 2 3   arl add 0 a1   System out println  Arraylist contains   arl get 0      It gives the output  Arraylist contains  I 3e25a5  Now my questions are    How to display the correct value i e  1 2 3  How can I access the single element of array a1 i e  if I want to know the value at a1 1

User · Answer

Integer is wrapper class and int is primitive data type Always prefer using Integer in ArrayList

User · Answer

You have to use  lt Integer gt  instead of  lt int gt    int a1      1 2 3   ArrayList lt Integer gt  arl new ArrayList lt Integer gt     for int i   a1        arl add i               System out println  Arraylist contains     arl get 0

User · Answer

The setup       List lt int   gt  intArrays new ArrayList lt  gt         int anExample    1 2 3       intArrays add anExample     To retrieve a single int   array in the ArrayList by index       int   anIntArray   intArrays get 0      0  is the index       iterate the retrieved array an print the individual elements     for  int aNumber   anIntArray              System out println  Arraylist contains     aNumber            To retrieve all int   arrays in the ArrayList         iterate the ArrayList  get and print the elements of each int   array       for int   anIntArray intArrays             iterate the retrieved array an print the individual elements        for  int aNumber   anIntArray               System out println  Arraylist contains     aNumber                Output formatting can be performed based on this logic  Goodluck

User · Answer

For the more inexperienced  I have decided to add an example to demonstrate how to input and output an ArrayList of Integer arrays based on this question here      ArrayList lt Integer   gt  arrayList   new ArrayList lt Integer   gt         while n  gt  0                int d   scan nextInt           Integer temp     new Integer d           for  int i   0   i  lt  d   i                          int t   scan nextInt                temp i  Integer valueOf t                     arrayList add temp           n--         n is the size of the ArrayList that has been taken as a user input  amp  d is the size        of each individual array          to print something  out from this ArrayList  we take in two         values index and index1 which is the number of the line we want and         and the position of the element within that line  since the question        followed a 1-based numbering scheme  I did not change it here       System out println Integer valueOf arrayList get index-1  index1-1      Thanks to this answer on this question here  I got the correct answer  I believe this satisfactorily answers OP s question  albeit a little late and can serve as an explanation for those with less experience

User · Answer

java util Arrays toString   converts Java arrays to a string   System out println  Arraylist contains   Arrays toString arl get 0

User · Answer

In java  an array is an object  Therefore the call to arl get 0  returns a primitive int   object which appears as ascii in your call to System out    The answer to your first question is therefore  System out println  Arraylist contains   Arrays toString  arl get  0          If you re looking for particular elements  the returned int   object must be referenced as such  The answer to your second question would be something like      int   contentFromList   arl get 0       for  int i   0  i  lt  contentFromList length  i              int j   contentFromList i           System out println  Value at index -   i   is    j

User · Answer

Use Arrays toString  arl get 0     arl get 0  1

User · Answer

ArrayList lt Integer gt  list   new ArrayList lt  gt     int number  total   0   for int i   0  i  lt   list size    i         System out println  Enter number      i   1      or enter -1 to end          number   input nextInt         list add number        if number    -1           list remove list size   - 1           break          System out println list toString      for int i  list       System out print i              total   i    System out println    System out println  The sum of the array content is      total

User · Answer

Everyone is right  You can t print an int   object out directly  but there s also no need to not use an ArrayList of integer arrays   Using   Arrays toString arl get 0     means splitting the String object into a substring if you want to insert anything in between  such as commas   Here s what I think amv was looking for from an int array viewpoint    System out println  Arraylist contains           arl get 0  0                arl get 0  1                arl get 0  2      This answer is a little late for amv but still may be useful to others

User · Answer

First of all  for initializing a container you cannot use a primitive type  i e  int  you can use int   but as you want just an array of integers  I see no use in that   Instead  you should use Integer  as follows   ArrayList lt Integer gt  arl   new ArrayList lt Integer gt       For adding elements  just use the add function   arl add 1     arl add 22   arl add -2     Last  but not least  for printing the ArrayList you may use the build-in functionality of toString     System out println  Arraylist contains      arl toString         If you want to access the i element  where i is an index from 0 to the length of the array-1  you can do a    int i   0     Index 0 is of the first element System out println  The first element is      arl get i      I suggest reading first on Java Containers  before starting to work with them

User · Answer

More simple than that   List lt Integer gt  arrayIntegers   new ArrayList lt  gt  Arrays asList 1 2 3     arrayIntegers get 1     In the first line you create the object and in the constructor you pass an array parameter to List   In the second line you have all the methods of the List  class   get

[java] ArrayList of int array in java

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