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Anthonyx82
Asked: 2022-10-20 02:18:55 +0800 CST 2022-10-20 02:18:55 +0800 CST

Alternative to if in java 2.0

  • 772

Hello good, previously I needed to make a code like this:

#Entrada (numero entero del 1 al 10)
#Salida en texto (se ha introducido el numero cinco

Example

#Introduzco el numero 6
#Haz introducido el numero "seis"

I discovered thanks to two users here that it could be done thanks to an array and a number that printed the position of the array that interested me thanks to the following code:

public static void main(String[]args) {
        try (Scanner sc = new Scanner(System.in)) {
            ArrayList<String> numeros = new ArrayList<String>();
            numeros.add("Cero");
            numeros.add("Uno");
            numeros.add("Dos");
            numeros.add("Tres");
            numeros.add("Cuatro");
            numeros.add("Cinco");
            numeros.add("Seis");
            numeros.add("Siete");
            numeros.add("Ocho");
            numeros.add("Nueve");
            numeros.add("Diez");
            int intro;
            System.out.println("Introdue un numero del 1 al 10 para saber como se escribe");
            intro = sc.nextInt();
            if (intro > 10 && intro < 0){
                System.out.println("Numero incorrecto");
            }
            else{
                System.out.println(numeros.get(intro));
            }
        }
    }

Now comes my problem and it is that I wanted to know if I can do the same thing but the other way around is to say that the user enters the number in letters and the program returns the whole number I mean:

#Introduzco el numero "seis"
#Haz introducido el numero 6

His thing would be not to do it with either if or switch unless there is no simpler alternative. Thanks in advance

java

2 Answers

  1. Best Answer

    To get the name of a number

    You can define a string array with each value and then search by index:

    String[] nameOfNumbers = {
        "cero", "uno", "dos", "tres", "cuatro", "cinco", 
        "seis", "siete", "ocho", "nueve", "diez"
    };

    And then:

    intro = sc.nextInt();
try 
{
    System.out.println(nameOfNumbers[intro]);
}
catch(ArrayIndexOutOfBoundsException exception) 
{
    System.out.println("Número incorrecto");
}

    The block try ... catchwill allow your code to work by substituting each existing value in the nameOfNumbers. So, if you wanted the , too once, you add it in its proper position, at the end of the current nameOfNumbers. With him try ... catchyou don't need this: if (intro > 10 && intro < 0){but if it's a requirement of an exercise whatever, remove the block try ... catch.

    It should be said that this works because in Java, as in almost all languages, arrays start at index 0, so in nameOfNumberswe simply assign the name that would correspond to each numerical value. For more demanding programs, in which you need to convert any number to letters or vice versa, there are libraries that already do it 1 , taking into account aspects as changing as the language.

    To get the number from the letter

    You can create a HashMapthat associates each numeric value with a name (this in reverse would help you to solve what was stated in the first case).

    Map<String,Integer> map = new HashMap<>();
    map.put("cero",0);
    map.put("uno",1);
    map.put("dos",2);
    map.put("tres",3);
    map.put("cuatro",4);
    map.put("cinco",5);
    map.put("seis",6);
    map.put("siete",7);        
    map.put("ocho",8);
    map.put("nueve",9);
    map.put("diez",10);

    Then, you would look up the value associated with that key. For example:

    String introS="tres";
if(map.containsKey(introS))
{
    System.out.printf("En letras %d equivale a %s",map.get(introS),introS);
}
else
{
    System.out.println("Número incorrecto");
}

    Departure:

    En letras 3 equivale a tres

    Grades

    1. See for example the following:

    • 2

  2. A somewhat cleaner alternative is to use the Java 8+ Optional.

    Something like that:

    private Optional<Integer> getValueFromNumberString(String theNumber) {
    Map<String,Integer> map = new HashMap<>();
    map.put("cero", 0);
    map.put("uno", 1);
    map.put("dos", 2);
    map.put("tres", 3);
    map.put("cuatro", 4);
    map.put("cinco", 5);
    map.put("seis", 6);
    map.put("siete",7);
    map.put("ocho", 8);
    map.put("nueve", 9);
    map.put("diez", 10);

    return Optional.ofNullable(map.get(theNumber.toLowerCase()));
}

private String stringToNumberMessage(int theValue, String theNumber) {
    return theNumber + " in numeric form is " + theValue;
}

private String errorMessage() {
    return "Number string not found";
}

    And then, to invoke it, for example with the "four":

        String theNumberinLetters = "cuatro";
    System.out.println(getValueFromNumberString(theNumberinLetters)
        .map((i) -> stringToNumberMessage(i, theNumberinLetters))
        .orElseGet(this::errorMessage));

    A good guide here: https://dzone.com/articles/using-optional-correctly-is-not-optional

    • 0