CODEWITHSUNDEEP
java
Connor Humphries
Connor Humphries

Reputation: 23

Converting a letter to a number using a switch statement within a method - Java

I am trying to convert a roman numeral entered by the user into the correct value it represents.

My code:

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    System.out.print("Please enter the a roman numberal (I, V, X, L, C, D, M): ");
    String romanNumeral = in.nextLine();
    romanNumeralToInt(romanNumeral);
    System.out.println(romanNumeral);

}

public static int romanNumeralToInt(String romanNumeral) {
    switch (romanNumeral) {
    case "I":
        romanNumeral = 1;
        break;
    case "V":
        romanNumeral = 5;
        break;
    case "X":
        romanNumeral = 10;
        break;
    case "L":
        romanNumeral = 50;
        break;
    case "C":
        romanNumeral = 100;
        break;
    case "D":
        romanNumeral = 500;
        break;
    case "M":
        romanNumeral = 1000;
        break;
    }
}

I have tried numerous different ways of converting this. I tried to use char instead of string but the same problem came up.

I have also tried to use the line:

public static int romanNumeralToInt(String romanNumeral, int romanDecimal) {

and then using:

    case "I":
        romanDecimal = 1;
        break;

Which didn't work. I have tried printing out just System.out.println(romanNumeral) which also didn't work. I am kinda stuck and don't really understand how I go about returning a value as an integer when it was inputted as a string in a method

Upvotes: 1

Views: 1195

Answers (4)

AS2
AS2

Reputation: 89

You don't need to convert it you can simply return integer values like this:

public static void main(String[] args) {

    Scanner in = new Scanner(System.in);
    System.out.print("Please enter the a roman numberal (I, V, X, L, C, D, M): ");
    String romanNumeral = in.nextLine();
    romanNumeralToInt(romanNumeral);
    System.out.println(romanNumeral);

}

public static int romanNumeralToInt(String romanNumeral) {

    switch (romanNumeral) {
    case "I":
        return 1;
    case "V":
        return 5;
    case "X":
        return 10;
    case "L":
        return 50;
    case "C":
        return 100;
    case "D":
        return 500;
    case "M":
        return 1000;
    }
}

Upvotes: 1

Shivam Puri
Shivam Puri

Reputation: 1666

The current code you have pasted here does not seem right. No return written for the method, even though it clearly has an int return type. Was it compiled correctly?

Anyhow, change your method to have a return;

public static int romanNumeralToInt(String romanNumeral) {
    int intNumeral = 0;
    switch (romanNumeral) {
    case "I":
        intNumeral = 1;
        break;
    case "V":
        intNumeral = 5;
        break;
    case "X":
        intNumeral = 10;
        break;
    case "L":
        intNumeral = 50;
        break;
    case "C":
        intNumeral = 100;
        break;
    case "D":
        intNumeral = 500;
        break;
    case "M":
        intNumeral = 1000;
        break;
    }
    return intNumeral;
}

Then in the main method let the methods return variable be stored in an int type variable;

int intNumericConverted = romanNumeralToInt(romanNumeral);

Upvotes: 1

Max Alexander Hanna
Max Alexander Hanna

Reputation: 3871

Welcome to stackoverflow! My first thought is that you're not using your return value from your function, to do this, add a return statement instead of a break statement. it should read

 case "I":
        return 1;

then return from it:

System.out.println(romanNumeralToInt("I"));

Or , simply make sure youre assigning strings instead of integers inside your function :

 case "I":
        romanNumeral = "1";
        break;

but to accomplish this, you'll need to change the scope of romanNumeral to global if im not mistaken and this wouldnt be the best practice, in my opinion

Upvotes: 1

Sheri
Sheri

Reputation: 1413

Integer class has static method toString() - you can use it:

int i = 1234;
String str = Integer.toString(i);

Returns a String object representing the specified integer. The argument is converted to signed decimal representation and returned as a string, exactly as if the argument and radix 10 were given as arguments to the toString(int, int) method.

Upvotes: 0

Related Questions