Not sure about the input type and if condition

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Hey guys, do you know why the input is in String, instead of int? 1) Does it need to be a String here (or could the provide an int, too), or why did Codegym give us Strings instead of ints as input? 2) Is it bad practice to use an if, to check if the number is positive here? 2.1) Why don't we need to check that in this case? Thank you for your time and efforts, and have a great Sunday ahead! :)

A two-dimensional array is an array of arrays. That is, an array where each cell holds a reference to an array.
But it is much easier to think of it as a table that has a number of rows (first dimension) and a number of columns (second dimension).
In this task, we will create such an array.

Implement the main(String[]) method, which:

Reads from the console a number R — the number of rows in the array. The number must be greater than 0.
Then it reads R numbers from the console (any numbers greater than 0).
Then it initializes the two-dimensional multiArray array using:
- the number of rows R;
- row arrays whose size corresponds to the numbers entered in step 2 (in the order they are entered).

Example:

The number 5 is entered.
The numbers 1, 7, 5, 9, and 3 are entered.
And we get an array like this:
[] [][][][][][][] [][][][][] [][][][][][][][][] [][][]

Requirements:

The Solution class should have an uninitialized public static int[][] multiArray variable.
The main(String[]) method must read N+1 numbers from the keyboard.
In the main(String[]) method, initialize and fill the two-dimensional array as outlined in the task conditions.

package en.codegym.task.pro.task05.task0511; import java.util.Scanner; /* Creating a two-dimensional array */ public class Solution { public static int[][] multiArray; public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int r = Integer.parseInt(scanner.nextLine()); if (r > 0) { multiArray = new int[r][]; for (int i = 0; i < r; i++) { int newI = Integer.parseInt(scanner.nextLine()); multiArray[i] = new int[newI]; } } } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

5 December 2022, 03:54

useful

Keyboard input is always as a String in Java. To be used as a number the computer needs it specifically converted to an appropriate type (an int in this case). Validating data is never bad practice. I personally try to indent my code as little as possible. Instead of:

if (r > 0) {
   // code
}

I would do:

if (r < 1) return;
// code

It looks way better than something like this:

if () {
   // code
   while () {
      if() {
         //code
      } else {
         // code
         if() {
            switch() {
                case :
                   // code
                case:
                   //code
             }
         }
      }
   }
}

Chrizzly

Level 12 , Earth, Germany

5 December 2022, 08:40

Thank you for the tips! If it's always a String, why do we have the nextInt() method, and why doesn't it work here? I thought, if the user types in a number and I use nextInt(), it's getting converted to an int with this method. So, why can't we use it in this case? I understand that everything typed into console is a String, but then I don't understand why there is this method nextInt(), while we can't make use of it here.

Thomas

Level 41 , Bayreuth, Germany

5 December 2022, 09:07

solution

That solution is not the only way to solve this task. There are various possibilities. You also could have used the simpler BufferedReader class. That doesn't have build in conversion in the first place. And it's these methods like nextInt() that can lead to confusion. nextInt() does not consume a whole line and converts it to an int. It just takes part of the line till it finds a delimiter. By default the delimiter is set to whitespace. So if you eg. enter 1 2<ENTER>, then nextInt() will give you a 1 while integer.parseInt("1 2") would throw an exception and tell a possible user right away that the program isn't working this way. And that has been done by the writer of the above code as well. I don't think Codegym tests in this way so you can go the nextInt() way, too. But keep in mind that the Scanner class has some more pitfalls. Just look at this. A string of two lines. The first line is an int, the second a String. If you read it with nextInt(), nextLine() then the line magically disappeares (of course not, it's what I told you above about delimiters ;) )

Scanner console = new Scanner("1\nhello world");
System.out.println(console.nextInt());
System.out.println(console.nextLine());

Oh, and if you replace the String in the above example with a System.in, then the program lets you enter the number but does not ask you for the string anymore. As said... be aware of the pitfalls and know the classes you're using.