Initializing a variable Java

Question

I'm writing a simple program with a do while loop and switch, which cna accept a mathematical operation and execute it for given 2 numbers.

The problem I have is, why should I initialize the result produced by the operation to zero at the beginning.

If I don't make ans=0, it gives me errors. If the given conditions are not met, some code parts are not executed and I don't need "ans".

package q3;
import java.util.Scanner;

public class Q3 {
public static void main(String[] args) {

    Scanner input = new Scanner(System.in);
    char operator;
    float no1, no2, ans=0; // <-------------- Why should I initialize ans

    do {
        System.out.println(" Mathematical Operations to be performed :");
        System.out.println("\t * Multiplication\n\t / Division\n\t + Addition\n\t - Subtraction");
        System.out.println("Press any other character to exit");
        System.out.print("Mathematical Operator : ");
        operator = input.next().charAt(0);

        if (operator == '*' || operator == '/' || operator == '+' || operator == '-') {
            System.out.print("Number 1: ");
            no1 = input.nextFloat();
            System.out.print("Number 2: ");
            no2 = input.nextFloat();

            switch (operator) {
                case '*':
                    ans = no1 * no2;
                    break;
                case '/':
                    ans = no1 / no2;
                    break;
                case '+':
                    ans = no1 + no2;
                    break;
                case '-':
                    ans = no1 - no2;
                    break;
            }

            System.out.println("The answer of " + no1 + operator + no2 + " = " + ans);

        }
    } while (operator == '*' || operator == '/' || operator == '+' || operator == '-');

}

}

Answer 1

Java requires that all local variables are initialised before they are used.

In your print line, you read the value of abs, but not all control paths set a value to it. (Even though you think you've covered all possibilities of your switch given the outer if, the compiler will not see things that way: some other thread could modify operator).

So your IDE / compiler is suggesting that you initialise it at the point of declaration.

Answer 2

This is because if no case evaluates to true, the value of ans will not be set. So you cannot use it.

You can overcome this by adding a default case, and setting the value of ans as 0 in that.

Answer 3

You should initialize ans=0; because you didn't have a default value for ans, for this you need to initialized it.
But if you add the defualt value you don't need to initialize it like this:

    ...
    case '-':
    ans = no1 - no2;
      break;
    default :
    ans = someValue; 
     break; 

Answer 4

Well, it could be that none of the case statements apply and as a result ans would still be un-initialized. And since local variables have to be initialised before they are being used, you get this error.

Answer 5

If you didnt initialize it, you ans will have a garbage value at first.

It is not compulsory to initialize it.

But your program will be a better program if you initialize it.