Javascript Prime Number Function Console Issue

Question

Apologies if this has been covered somewhere. I've searched through a lot of prime number function answers on here but I cannot seem to find the answer to my issue. I have cobbled together a function to check if a number is prime, and then I'm running a for loop for numbers under 100 and printing out primes. My code does not print primes. It seems to start printing primes, but then throws a 9 in there, and it gets worse from there.

I found code that is seemingly identical to mine that successfully prints primes to the console. Below is my code; where is my error? For the record I am using jsfiddle, and I am very new at this.

function isPrime(number) {
  if (number < 2) {
    return false;
  }
  for (var i = 2; i < number; i++) { 
    if (number % i == 0) { 
      return false;
    } else {
      return true;
      }
  }
};

for (var i = 0; i < 100; i++) {
  if(isPrime(i)) {
   console.log(i);
  }
}

One "speed" improvement for this slow method of determining prime ... only need to loop up to `Math.sqrt(number);` - not an issue with such small numbers though — Jaromanda X, Dec 23 '15 at 14:08

lex82 · Answer 1 · 2015-12-23T14:11:23.253

0

You return true as soon as number is not divisible by i. Move your return true to the end:

function isPrime(number) {
    if (number < 2) return false;

    for (var i = 2; i < number; i++) { 
        if (number % i == 0) return false;
    }
    return true;
}

You only want to return true when you didn't find an i that divides number.

edited Dec 23 '15 at 14:11

answered Dec 23 '15 at 13:59

lex82

11,173
2
44
69

Thank you! That makes it work! Also, good point Jaromanda X, I will fix that as well. – Frances Dec 23 '15 at 14:03
1

sorry, wasn't thinking straight, comment removed – Jaromanda X Dec 23 '15 at 14:04
@jaromanda-x Thanks. You may also remove the downvote if it is from you ;-) – lex82 Dec 23 '15 at 14:07
@Frances you're welcome to accept the answer if it worked. – lex82 Dec 24 '15 at 00:51

Костя Самсин · Answer 2 · 2015-12-23T14:19:17.040

0

function isPrime(number) {
  var result = true;
  if (number < 2) {
    result = false;
  }
  for (var i = 2; i < number; i++) { 
    if (number % i == 0) { 
      result = false;
    }
  }
  return result;
};

for (var i = 0; i < 100; i++) {
  if(isPrime(i)) {
   console.log(i);
  }
}

edited Dec 23 '15 at 14:19

answered Dec 23 '15 at 14:06

Костя Самсин

57
2

yes you've corrected it; but you made it much much worse, speed-wise, when you replaced `return false` with `result=false`. – Will Ness Dec 23 '15 at 16:40

score 0 · Answer 3 · answered Dec 23 '15 at 14:51

Your issue is in

if (number % i == 0) { 
  return false;
} else {
  return true;
}

As you return something in every case (true or false), your algorithm exits on every first cycle.

You should not return anything in the else statement :

if (number % i == 0) { 
  return false;
} else {
  // continue the for statement.
}

And get the function to return true after the for.

For the joke, a much faster algorithm would be :

function isPrime(number) {
    var primes=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];
    if (primes.indexOf(number) > -1) {
        return true;
    } else {
        return false;
    }
};

Umar Sharjeel · Answer 4 · 2015-12-23T14:08:14.963

-2

function isPrime(number) {
var check=true;
  if (number < 2) {
    return false;
  }
  for (var i = 2; i < number; i++) { 
    if (number % i == 0) { 
      check=false;
    } 
  }
   return check;
};

for (var i = 0; i < 100; i++) {
  if(isPrime(i)) {
   console.log(i);
  }
}

edited Dec 23 '15 at 14:08

answered Dec 23 '15 at 14:05

Umar Sharjeel

108
10

No! You may not set check back to true once it has been set to false. – lex82 Dec 23 '15 at 14:06

Javascript Prime Number Function Console Issue

4 Answers4