Eclipse - Java, casting from double to int

Question

I just started using Eclipse and I'm quite new to Java. I have a block of code which won't compile because the random number I generated won't cast in to an integer. The error message reads "This method must return a result of type int" on line two in the following block of code.

public class passwordHelper {
    public int rNum(String nums, String lets){


        int z;


        if(nums == "yes" && lets == "yes"){
            z = (int)Math.random()*36;
            return z;

        } else if(nums == "yes"){
            z = (int)Math.random()*10;
            return z;   

        } else if(lets == "yes"){
            z = (int)Math.random()*26 + 10;
            return z;

        } else {
            System.out.println("Sorry, you need either letters or numbers in your password.");
        }
    }
}

As you can see I am using the "(int)" function to cast my number to an integer but it sends me the same error message, I have also tried using other methods of casting, such as "Math.floor()", "Math.round()" and combinations of all the three. And if anyone wants to know this is part of a code to generate a random string of numbers and letters for the user.

//Thanks for any help

String comparison is done by equals method. try to change nums == "yes" to nums.equals("yes") and the others — Devrim, Aug 26 '14 at 22:20
I suppose you have reasons but you should really be using boolean here instead of Strings. — Nyamiou The Galeanthrope, Aug 26 '14 at 23:08

score 2 · Accepted Answer · edited Aug 26 '14 at 22:42

You have to return a value in all branches, there are 3 possibilities:

1) Return an int in the else branch:

public int rNum(String nums, String lets){

    int z;

    if(nums == "yes" && lets == "yes"){
        z = (int)Math.random()*36;
        return z;

    } else if(nums == "yes"){
        z = (int)Math.random()*10;
        return z;   

    } else if(lets == "yes"){
        z = (int)Math.random()*26 + 10;
        return z;

    } else {
        System.out.println("Sorry, you need either letters or numbers in your password.");
            return -1;
    }
}

2) Throw an unchecked exception

public int rNum(String nums, String lets){

    int z;

    if(nums == "yes" && lets == "yes"){
        z = (int)Math.random()*36;
        return z;

    } else if(nums == "yes"){
        z = (int)Math.random()*10;
        return z;   

    } else if(lets == "yes"){
        z = (int)Math.random()*26 + 10;
        return z;

    } else {
        System.out.println("Sorry, you need either letters or numbers in your password.");
        throw new RuntimeException("Sorry, you need either letters or numbers in your password.");
    }
}

3) Throw a checked exception

public int rNum(String nums, String lets) throws Exception {

    int z;

    if(nums == "yes" && lets == "yes"){
        z = (int)Math.random()*36;
        return z;

    } else if(nums == "yes"){
        z = (int)Math.random()*10;
        return z;   

    } else if(lets == "yes"){
        z = (int)Math.random()*26 + 10;
        return z;

    } else {
        System.out.println("Sorry, you need either letters or numbers in your password.");
        throw new Exception("Sorry, you need either letters or numbers in your password.");
    }
}

Why? Is there a reason? Performance? IMHO it gets compiled into the same byte code anyways... — Romeo Kienzler, Sep 09 '14 at 05:15
"==" does not compile into the same byte code as .equals() http://stackoverflow.com/questions/1643067/whats-the-difference-between-equals-and — JustinKSU, Sep 09 '14 at 14:33

score 1 · Answer 2 · answered Aug 26 '14 at 22:22

1

your method must return an int value but at your last else block you are not returning anything. return a default value there, for instance return 0 or throw an exception which says invalid input.

And also;

To compare two strings you must use its equals method.

change nums == "yes" to "yes".equals(nums)

and

lets == "yes" to "yes".equals(lets)

answered Aug 26 '14 at 22:22

Devrim

15,345
4
66
74

While definetly true, and will come up later - this is not the question, the comment was enough. – amit Aug 26 '14 at 22:25
eagean will not be notified until you add "@name" in your comment. Usually you are always notified for your own question and answer when someone comment you, but not vice-versa. – afzalex Aug 26 '14 at 23:28
Great thanks and I actually ended up hitting the second issue you brought up so you sorted that for me as well, can't thank you enough. – Ivar Eriksson Aug 27 '14 at 13:02

score 1 · Answer 3 · edited Aug 26 '14 at 23:04

Try this:

public int rNum(String nums, String lets){

    int z;
    Random random = new Random();

    if(nums.equals("yes") && lets.equals("yes")){
        z = random.nextInt(36);// generates a random number from 0 to 36
        return z;

    } else if(nums.equals("yes")){
        z = random.nextInt(10);//generates a random number from 0 to 10
        return z;   

    } else if(lets.equals("yes")){
        z = random.nextInt(26 + 10);//generates a random number from 0 to 36
        return z;

    } else {
        System.out.println("Sorry, you need either letters or numbers in your password.");
            return -1;
    }
}

ROMANIA_engineer · Answer 4 · 2014-08-26T22:25:41.990

0

Math.random() returns a number between 0 and 1. If you cast it to int, you'll obtain 0. You can multiply the double, and then add cast using parantheses.

Alternative: You can use:

Random rand = new Random();

and then

rand.nextInt(...);

P.S.:

add a return statement (at the end)
change String comparison to use "equals"

edited Aug 26 '14 at 22:25

answered Aug 26 '14 at 22:20

ROMANIA_engineer

54,432
29
203
199

score 0 · Answer 5 · answered Aug 26 '14 at 23:19

String is a reference type and you cannot compare a reference type with comparative operators. i.e. You cannot use == here.
To compare Strings you can use equals() or equalsIgnoreCase().
equalsIgnoreCase() doesnot consider casing while comparing two strings, and I prefer this to use here. But here I modified your code by replacing == with equals(). If you do not want to consider casing i.e. want true for yes as well as YES here then use equalsIgnoreCase() instead.

int z;
if(nums.equals("yes") && lets.equals("yes")){
    z = (int)Math.random()*36;
    return z;
} else if(nums.equals("yes")){
    z = (int)Math.random()*10;
    return z;   
} else if(lets.equals("yes")){
    z = (int)Math.random()*26 + 10;
    return z;
} else {
    System.out.println("Sorry, you need either letters or numbers in your password.");
}

Eclipse - Java, casting from double to int

5 Answers5