Return a boolean value using if/else in java

Question

I am trying to write a code to check if a number is prime and return true if it is a prime number using if/else statement.

I am giving the code below, I have written. It is always showing an exception "missing return statement".

class Main
{
    static boolean isPrime(int x)
    {
        for(int i = 2;i <= x/2;++i)
        {     
            if(x%i == 0)
            {
                return false;
            }
            else
            {
                return true;
            }   
        }
    }

    public static void main (String args[])
    {              
        boolean prime = isPrime(11);
        System.out.println(prime);           
    }
}

There are odd cases where the for-loop doesn't run. The compiler wants a return at the end. — NomadMaker, Jan 30 '21 at 14:04
Hint: `if(condition) { return true; } else { return false; }` is same as `return condition;`. So your loop looks like `for(int i = 2;i <= x/2;++i) { return x%i == 0; }` which defeats the purpose of loop as you are always returning some value in the first iteration, which prevents other iterations). Try to think more about each of your if/else branches. Are you really able to return correct value in each of them? Maybe some value can be returned only after all iterations will check their conditions? — Pshemo, Jan 30 '21 at 14:06

Roberto Caboni · Answer 1 · 2021-01-30T18:05:56.800

The problem in your code is that you are returning true the first time the modulus is not 0. That's wrong, because you have to make sure that x is not divisible to any of the numbers in range [2, x/2].

Modify your method as follows:

static boolean isPrime(int x)
{
    if(x <= 1)
    {
        return false;
    }

    boolean ret = true;

    for(int i = 2;i <= x/2;++i)
    { 
         if(x%i == 0)
         {
              ret = false;
              break;
         }           
     }

     return  ret;
 }

In this way you break as soon as you are sure the number is not prime. If it's prime it will complete the loop without finding non-zero remainders.

The x<=1 scenarios is the cause of the warning you are experincing, as with those inputs your method exits without encountering a return statement. I just checked for this condition at the beginning of the function.

Finding prime numbers

I just suggested the fix to your isPrime implementation, which is the simpliest way to find prime numbers. Anyway there are smarter ways to find them:

The Siege of Eratosthenes method
You could check only odd numbers, saving half the time
You could check if x belongs to the 6*x±1 set. In fact all prime numbers from number 5 follow this rule

score 2 · Answer 2 · answered Jan 30 '21 at 13:47

2

You are not taking into account that the number "x" passed into the method can be zero. If that is the case (x == 0) you loop will not execute and there is a path through the method that does not return TRUE or FALSE.

answered Jan 30 '21 at 13:47

Peter Knall

86
5

Nowhere Man · Answer 3 · 2021-01-30T15:14:39.267

As explained before, in the existing implementation it is possible that loop is not executed and no return statement is provided after the loop.

Better search for prime numbers would be:

Check for 0 and 1 to return false for them
Check for even numbers: all even numbers except 2 are composite
Check only odd numbers for primality
Check limited range of numbers using "inverse" square root condition: i * i <= n
(Optionally) Exclude negative numbers

Example implementation:

static boolean isPrime(int x) {
    int n = Math.abs(x); // exclude negatives
    
    if (n < 2) return false;        // exclude 0, 1
    if (n % 2 == 0) return n == 2;  // exclude even except 2
     
    for (int i = 3; i * i <= n; i += 2) {  // check only odd numbers
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

Faster implementation using 6n ± 1 rule (thanks Roberto Caboni) could look like this (starting from 5 (6n - 1) and using variable steps 2 and 4):

static boolean isPrimeFaster(int x) {
    int n = Math.abs(x);
    
    if (n < 2) return false;       // exclude 0, 1
    if (n % 2 == 0) return n == 2; // exclude even except 2
    if (n % 3 == 0) return n == 3; // exclude multiples of 3

    for (int i = 5, j = 2; i * i <= n; i += j, j = j == 2 ? 4 : 2) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

Aalexander · Answer 4 · 2021-01-31T17:37:00.253

Use a boolean variable and set it true when the number is divisible. Also check if the number is less than 2 because one and zero are not primes so return false then.

The problem in your code currently is that you immediately return false or true after the first iteration but you have to check as long as you found a divisor or you run up from 2 until the square root of x This is more accurate to run until the iterator is less or equal than the square root of the prime

If you are interested why to use the square root as running condition check out this answer

Why do we check up to the square root of a prime number to determine if it is prime?

public static void main(String[] args) {
    
System.out.println(isPrime(0));
System.out.println(isPrime(3));
System.out.println(isPrime(4));
System.out.println(isPrime(5));
System.out.println(isPrime(11));
}

public static boolean isPrime(int x) {
    boolean isPrime = true;; 
    
    if(x < 2) {
        return false;
    }
    for(int i = 2; i <= Math.sqrt(x); i++) {
        if (x % i == 0) {
            isPrime = false;
            break;
        } 
    }
    
    return isPrime;

 }

Generates the output

false
true
false
true
true

score 0 · Answer 5 · answered Jan 30 '21 at 14:48

You need this implementation

static boolean isPrime(int x)
{
    If(x==0||x==1) {
          return false;
     }
    for(int i = 2;i <= x/2;++i)
    {     
        if(x%i == 0)
        {
            return false;
        }
     }
     return true;
}

score 0 · Answer 6 · answered Jan 31 '21 at 16:49

The compiler is confused because both return statements are conditional. It isn't sure any one of them will happen. Your program will compile if you simply move the return true; outside the for loop.

class Main
{
    static boolean isPrime(int x)
    {
        for(int i = 2;i <= x/2;++i)
        {
            if(x%i == 0)
            {
                return false;
            }
        }
        return true;
    }

    public static void main (String args[])
    {
        boolean prime = isPrime(11);
        System.out.println(prime);
    }
}

Your program will not return a correct answer fo x <= 2. We can fix that by adding a few ifs at the beginning.

class Main
{
    static boolean isPrime(int x)
    {
        if (x < 2)
        {
            return false;
        }
        if (x == 2)
        {
            return true;
        }
        for(int i = 2;i <= x/2;++i)
        {
            if(x%i == 0)
            {
                return false;
            }
        }
        return true;
    }

    public static void main (String args[])
    {
        boolean prime = isPrime(11);
        System.out.println(prime);
    }
}

You can improve the performance of your program without too much of a change by limiting your search to sqrt(x) instead of x/2. And now that we are optimising, we might as well only check the odd numbers.

class Main
{
    static boolean isPrime(int x)
    {
        if (x < 2)
        {
            return false;
        }
        if (x == 2)
        {
            return true;
        }
        if (x % 2 == 0)
        {
            return false;
        }
        int limit = (int) Math.sqrt(x);
        for(int i = 3; i <= limit; i += 2)
        {
            if(x%i == 0)
            {
                return false;
            }
        }
        return true;
    }

    public static void main (String args[])
    {
        boolean prime = isPrime(11);
        System.out.println(prime);
    }
}

score -1 · Answer 7 · answered Jan 30 '21 at 13:47

-1

java compiler expects isPrime() always returns a boolean.
you can loop and when remaining become 0 return false. and after and loop return true.

static boolean isPrime(int x)
    {
        for(int i = 2;i <= x/2;++i)
            if(x%i == 0)
                return false;
        return true;
    }

answered Jan 30 '21 at 13:47

Ali Choopani

75
7

This code returns `true` for 0 and 1 which is not correct: 0 and 1 are [neither prime nor composite numbers](https://brilliant.org/wiki/is-0-prime/). – Nowhere Man Jan 30 '21 at 14:13
@AlexRudenko this is obvious. I just correct his algorithm. – Ali Choopani Jan 31 '21 at 07:16

score -3 · Answer 8 · answered Jan 30 '21 at 13:48

-3

class PrimeNumber {

   static boolean isPrime(int x)
       {

           boolean status = false;
            for(int i = 2;i <= x/2;++i)
            {

                if(x%i == 0)
                {
                  status = false;
                }
                else
                {
                 status = true;
                }
            }

         return status;
    }

   public static void main (String args[])
   {
       boolean prime = isPrime(11);
       System.out.println(prime);
   }

}

this will solve your problem

answered Jan 30 '21 at 13:48

Prasad Kadam

1
1

I have tried this but it is returning odd numbers too as a prime. – Gaurav Kumar Jan 30 '21 at 13:56
This code is not correct, it returns that the primes are 9, 15, 21, 25, 27, etc. and 2, 3 are NOT prime – Nowhere Man Jan 30 '21 at 14:03

Return a boolean value using if/else in java

8 Answers8

Finding prime numbers

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