Java homework help

Question

Hello

I have this assignment that I've tried. But when I enter 1 it should not give any output as 1 is > 0 and 1 is not even but I still get output as:

Enter a +ve number
1
You entered 1
I'd asked for a +ve number :)

.

 import java.util.Scanner;
 class Main {
    public static void main(String[] args) {
      Scanner input = new Scanner(System.in);
      System.out.println("Enter a +ve number");
      int number = input.nextInt();
      System.out.println("You entered "+number);
      if(number > 0)
         if(number %2 == 0)
             System.out.println("Number"+number+" is even and +ve");
         else
             System.out.println("I'd asked for a +ve number :)");
     }
 }

Answer 1

Your else actually belongs to the 2nd if not the 1st if as the indentation shows.

if(cond1)
 if(cond2) 
else   // this else belongs to 2nd if not 1st if.

is same as:

if(cond1) {
 if(cond2) {
 } else {
 }
}

This is because the Java grammar says that an else belongs to the closest unmatched if to which it can possibly belong.

If you want to match the else with first if you need to use parenthesis as:

if(cond1) {
 if(cond2) {
 } 
} else { 
}

Answer 2

Check that the code actually follows the logic it ought to - indentation won't help you with flow control, that's what curly brackets {} are for.

Answer 3

first you should try to debug (using eclipse maybe) add paranthesis after if(number > 0) statement to wrap inner if statement.

Answer 4

I get the following output when I enter 1:

Enter a +ve number
1
You entered 1
I'd asked for a +ve number :)

Which is reasonable as you first check

if (number > 0)

which is true for number == 1 and then you check

if (number % 2 == 0)

which is false for number == 1 thus the else branch is taken:

System.out.println("I'd asked for a +ve number :)")

Your code should probably look something like:

Scanner input = new Scanner(System.in);
System.out.println("Enter a +ve number");
int number = input.nextInt();
System.out.println("You entered " + number);
if (number > 0) {
    if (number % 2 == 0)
        System.out.println("Number" + number + " is even and +ve");
} else {
    System.out.println("I'd asked for a +ve number :)");
}

Answer 5

That would probably be because your ifs are interpreted like this:

if(number > 0)
    if(number %2 == 0)
        System.out.println("Number"+number+" is even and +ve");
    else
        System.out.println("I'd asked for a +ve number :)");

How's the computer supposed to know what you mean? You need to use brackets to make it unambiguous:

if(number > 0) {
    if(number %2 == 0) {
        System.out.println("Number"+number+" is even and +ve");
    }
} else {
    System.out.println("I'd asked for a +ve number :)");
}

Answer 6

I threw your code into Netbeans and had it auto format it for readability. Is this your intended logic flow? Because this is how the compiler sees your if statements alignment/nesting.

import java.util.Scanner;

public class Main
{


    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        System.out.println("Enter a +ve number");
        int number = input.nextInt();
        System.out.println("You entered " + number);

        if (number > 0)
        {
            if (number % 2 == 0)
            {
                System.out.println("Number" + number + " is even and +ve");
            } 
            else
            {
                System.out.println("I'd asked for a +ve number :)");
            }
        }
    }

}

Answer 7

1 %2 == 1 so it's clear that it enters on the else branch