CODEWITHSUNDEEP
javascriptif-statementreturnlogic
sleepsdumb
sleepsdumb

Reputation: 3

I cannot understand this specific syntax of the RETURN instruction within the IF argument

I am learning JS through some exercises. One of them is about writing a function that checks whether a number is prime or not. The correct answer to this exercise is this one:

const isPrime = num => {
  for(let i = 2; i < num; i++)
    if(num % i === 0) return false;
  return num > 1;
}

I'm accustomed to inserting return false; inside the if argument and the return num > 1;or return true; outside of it, right after the closing bracket that ends the if argument, as with the if statement I am usually telling to my function "if this condition fulfilled, return this" and right after having closed the if statement I'm telling it "...otherwise return that". I cannot understand why in this case both the return instructions have be inserted inside the argument, what's the logic behind that?

Upvotes: 0

Views: 44

Answers (1)

BEAGLE
BEAGLE

Reputation: 343

This

const isPrime = num => {
  for(let i = 2; i < num; i++)
    if(num % i === 0) return false;
  return num > 1;
}

is equivalent to

const isPrime = num => {
    for(let i = 2; i < num; i++){
        if(num % i === 0){
            return false;
        }
    }
    return num > 1;
}

Explanation

Blocks of codes are usually enclosed in curly braces { code }. So when some part of code needs to be executed inside a for loop or after if condition we enclose it in curly braces. However, if there is only one statement that needs to be executed after if or inside for loop, we don't have to use braces. The code considers the immediate next line after if condition/for loop belongs to it.

//if
if(condtion){
    console.log("hello world");
}
//for
for(let i=0;i<num;i++){
    console.log(i);
}

//also equals to 
if(condition)
    console.log("hello world");
//for
for(let i=0;i<num;i++)
    console.log(i);

Upvotes: 2

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