I found this code that checks the argument to see if it's a prime number and for the most part I got it until I reached the very end.
const isPrime = num => {
for(let i = 2; i < num; i++)
if(num % i === 0) return false;
return num > 1;
}
what does return num > 1 do?
You would think it could just return true at the end, as that means the for loop did not detect any divisor for num, which is what it means for a number to be prime.
The problem is that it would then also return true for when num is 1. But 1 is not a prime number (by definition), so that would be wrong.
This is the reason why it has return num > 1. This is short for:
if (num > 1) {
return true;
} else {
return false;
}
Or:
return num > 1 ? true : false;
But these are really antipatterns, because they translate a boolean expression in ... the same boolean value. You can just return the boolean value that the comparison results in, and that is what return num > 1 does.
num > 1 returns a Boolean value depending on value for num.
true if num is greater than 1false if num is less or equal to 1let num = 5;
console.log(num > 1); // Logs true
num = 0;
console.log(num > 1); // Logs false
num = 1;
console.log(num > 1); // Logs falseAs per the function definition the function is checking the input number is prime or not.
What is a prime number?
A prime number is a number which is divisible only with 1 and that number.
So here in the function the number is consecutively divided by numbers from 2 to that number - 1. If any of the modulus operations return zero, that number is not prime and it returns false.
If nothing return false, then the number should be greater than 1
Please note 1 is not a prime number.
That is why the last comparison is done. To check whether the number is greater than 1. If the number is greater than 1, the number will be prime.
Because if the number is not divisible by any number from 2 to number - 1, that number should be greater than 1 to be prime number.
isPrime() function, not what it means as a JS operation.This is because 1 is not prime and after the finishing the above loop if compiler reach on this level then number is definitely prime
numis greater than 1.numand1-trueif the number is greater than1,falseotherwise. I’d recommend brushing up on your fundamentals1a prime number? Is zero a prime number? math.stackexchange.com/questions/120/…i < num, what don't you understand aboutnum > 1?