使用接头时为什么我的功能不能正常工作？

Question

I am trying to write a function which should calculate all prime numbers up to an input parameter and return it. I am doing this for practice.

I wrote this function in a few ways but I was trying to find new ways to do this for more practice and better performance. The last thing I tried was the code below:

 function primes(num){ let s = []; // sieve for(let i = 2; i <= num; i++){ s.push(i); } for(let i = 0; i < s.length; i++) { for(let j = s[i]*s[i]; j <= num;) { //console.log(j); if(s.indexOf(j)!= -1){ s.splice(s.indexOf(j), 1, 0); } j+=s[i]; } } s = s.filter(a => a != 0); return s; } console.log(primes(10)); 

The problem is that when I run this in a browser it keeps calculating and won't stop and I don't know why.

Note: when I comment out the splice and uncomment console.log(j); everything works as expected and logs are the things they should be but with splice , the browser keep calculating and won't stop.

I am using the latest version of Chrome but I don't think that can have anything to do with the problem.

Answer 1

Your problem lies in this line:

s.splice(s.indexOf(j), 1, 0);

Splice function third argument contains elements to be added in place of the removed elements. Which means that instead of removing elements, you are swapping their values with 0's, which then freezes your j-loop. To fix it, simply omit third parameter.

 function primes(num){ let s = []; // sieve for(let i = 2; i <= num; i++){ s.push(i); } for(let i = 0; i < s.length; i++) { for(let j = s[i]*s[i]; j <= num;) { //console.log(j); if(s.indexOf(j)!= -1){ s.splice(s.indexOf(j), 1); } j+=s[i]; } } return s; } console.log(primes(10)); 

Answer 2

Your problem is in this loop:

for(let j = s[i]*s[i]; j <= num;)

This for loop is looping forever because j is always less than or equal to num in whatever case you're testing. It is very difficult to determine exactly when this code will start looping infinitely because you are modifying the list as you loop.

In effect though, the splice command will be called setting some portion of the indexes in s to 0 which means that j+=s[i] will no longer get you out of the loop.