Question about static int variables on the code.

Question about the task

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So, the last task (Odd or Even). I don't see the following int variables initialized to zero anywhere, but the program somehow works just fine without errors: public static int even; public static int odd; What am I missing? TIA!

Use the keyboard to enter a positive number. Determine the number of even digits and odd digits in the entered number.
If a number is divisible by 2 without a remainder (i.e. the remainder is zero), then it is even.
And we'll increase the even digit counter (static variable even) by 1.
Otherwise, the number is odd, so we'll increase the odd digit counter (static variable odd).
Display the following message: "Even: a Odd: b", where a is the number of even digits and b is the number of odd digits.

Example for 4445:
Even: 3 Odd: 1

Requirements:

The program must read data from the keyboard.
The main method should calculate how many even digits are in the entered number and then write this number to the variable even.
The main method should calculate how many odd digits are in the entered number and then write this number to the variable odd.
The program should display text on the screen.
The displayed text must match the task conditions.

package com.codegym.task.task06.task0606; import java.io.*; /* Even and odd digits */ public class Solution { public static int even; public static int odd; public static void main(String[] args) throws IOException { //write your code here BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String input = reader.readLine(); //even = 0; //odd = 0; for (int i = 0; i<input.length(); i++){ //System.out.println(input.charAt(i)); int x = input.charAt(i)-'0'; //System.out.println(x); if (x%2==0){ even++; } else{ odd++; } } System.out.println("Even: "+even+" Odd: "+odd); } }

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Deepak Chandra

Level 16 , Visakhapatnam, India

26 May 2019, 14:34

solution

Static variables are class variables , that means they belong to the class and don't need to be initialized immediately after declared and such variables can be used any where inside a class and in instances of classes as well (depending on the access modifier you use) by calling ClassName.StaticVariableName as you know. So as you read the underlined part above it's not mandatory to initialize them while declaring because they already store default value for example in above task the static variables even and odd store 0 which is the default value for an int variable if it is not declared . (Note : - this won't be true for every static variable , it depends on type of the static variable you are declaring , if the type of the static variable you are declaring does not have a default value or if its an object then it needs to be initialized immediately.) To check comment out every thing inside main method and just keep the print out statement without changing anything see the out put you will know yourself.

John Michael Montuya

Level 10 , Cebu City, Philippines

26 May 2019, 21:42

That makes more sense now and that answers my question directly, so thanks man! Thank you so much!

ThomasLC

Level 16 , Singapore, Singapore

25 May 2019, 17:04

What is the Is the - '0' at line 26 supposed to do? If you're trying to convert a character to an int. Their unicode number gets converted instead so I.e. The character 4 gets converted to 52. Try to rework your algorithm logic

John Michael Montuya

Level 10 , Cebu City, Philippines

26 May 2019, 07:54

Sure, my point is that the variables odd and even are not initialized, or at least I think they are not because I can't see them being assigned a temporary value.

John Michael Montuya

Level 10 , Cebu City, Philippines

26 May 2019, 21:41

Also, to address your concern, my algorithm works perfectly fine and I have passed the task with only 1 attempt. Thank you.

Guadalupe Gagnon

Level 37 , Tampa, United States

26 May 2019, 22:44

chars can be converted to ints because chars are actually stored as numbers, which you can find their integer values on any ASCII table. I use https://www.asciitable.com/. Subtracting the char '0' from the char stored in the list at index i subtracts 48 (the ASCII value of '0') from whatever the ASCII value of that char is. So if the char is '3', which is equal to ASCII 51, then by subtracting '0' from it you would be left with ASCII 3, which would be the char for 'end of text'. If the char is 'U', subtracting '0' from it would leave you with 37, which is the char '%'. Here is some code to demonstrate:

public class Solution {
    public static void main(String[] args) throws Exception {
        String testChars = "U7mz'";
        char sutractChar = '0';

        System.out.println("Test program to subtract " + sutractChar + "(which is equivalent to ASCII " + (int)sutractChar +") from each char in: " + testChars);


        for(char x : testChars.toCharArray()){
            System.out.println("===================================================================");
            System.out.println("\tTesting char: " + x +" - ASCII value: " + (int)x);
            System.out.println("\tNew char value after subtracting 0: " + (char)(x-sutractChar));
            System.out.println("\tNew int value after subtracting 0: " + (x-sutractChar));
        }
    }

}

ThomasLC

Level 16 , Singapore, Singapore

27 May 2019, 04:50

Thanks Guadalupe. Basically don't use char in your equation because what you think are '3' and '4' are actually converted to a whole different number as the actual integer value for chars are based on their ascii map

Guadalupe Gagnon

Level 37 , Tampa, United States

27 May 2019, 05:31

useful

@ThomasLC In this example the Ascii ints for each char digit are even and odd in parallel with the actual numeral digit, so you can mod 2 on each char and get the correct output, and subtracting the char '0' from them still did not throw the output off because '0' is even. If you take an even or odd number and subtract it by an even number, the result will be even/odd depending on the first number, while subtracting two odd numbers will give you even numbers. I wouldn't suggest continue using this method though unless you specifically mean to, the OP had a good idea but kind of got lucky on the results.