Release of exceptions as a function of prime number

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Write a function that receives a number, throw an exception if the number is not positive and return true if it is prime, or false, otherwise.

My answer:

public static void main(String []args) throws Exception{
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    if(n <= 0){
        throw new Exception("Número não pode ser menor ou igual à zero!");
    }
    else{
        primo(n);
    }
}
public static void primo(int n){
    int contador = 0;

    for(int j = 1; j <= Math.sqrt(n); j++){
        if(n % j == 0){
            contador++;
            if(contador == 2){
                break;
            }
        }
    }
    if(contador == 1){
        System.out.println("Verdadeiro");
    }
    else{
        System.out.println("Falso");
    }
}

How can I rewrite my answer in the form of a single boolean function?

Ex: public static boolean f(int n)

java exception boolean
  • 1

    Normally I would use IllegalArgumentException. Use only Exception is generally a bad programming practice.

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2 answers

4

Normally I would not use exception for this, I think an abuse, it is good to make the function not print anything and return the result to print off it, so separates the responsibilities. I do not find suitable exception in many situations, launch Exception even though it’s too generic. But if you want to do it, this would be it:

import java.util.*;

class Ideone {
    public static void main(String []args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        try {
            System.out.println(primo(n) ? "Verdadeiro" : "Falso");
        } catch (IllegalArgumentException ex) {
            System.out.println(ex.getMessage());
        }
    }
    public static boolean primo(int n) throws IllegalArgumentException {
        if (n <= 0) throw new IllegalArgumentException("Número não pode ser menor ou igual à zero!");
        int contador = 0;
        for (int j = 1; j <= Math.sqrt(n); j++){
            if (n % j == 0) {
                contador++;
                if (contador == 2) return false;
            }
        }
        return true;
    }
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Read this: Why should we avoid returning error codes?. Look for more about the subject on the website.

3

It can even simplify a lot:

public static boolean primo( int n )
{
    int contador = 0;

    for( int j = 1; j <= Math.sqrt( n ); j++ )
        if( n % j == 0 )
            if( ++contador == 2 )
                return false;

    return true;
}

See working

  • From what I understand it seems he’s worried about the exception.

  • @bigown because I didn’t notice it like that. Let’s see what he says ;)

  • This question is from a list of college exercises. From what I understand, the teacher wants to write a single function that plays an exception if the number is negative (Illegalargumentexception) and returns true if the number is prime. But I realized that it is also necessary to check if the number received is zero.

  • @Caio_ignatz then the reply from the bigown answers your question.

Browser other questions tagged java exception boolean

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