Sieve of Eratosthenes | 易学教程

问题

I read up on the sieve of Eratosthenes while solving a question on Project Euler. I'm sure you guys know which question im talking about. So here's the thing. My code manages to show all the primes under 1 million correctly. However when i try the same implementation for 2 million it's giving me a segmentation fault... I have a certain idea of why the error is coming but don't know how to correct it... Here's the code for primes under 1 million.

#include<stdio.h>
int main(void)
{
   int i,k=2;
   int j;
   int n=1000000;
   int prime[2000000]={};
   for(i=0;i<n;i++) // initializes the prime number array
   {
      prime[i]=i;
   }
   for(i=2;i<n;i++) // Implementation of the Sieve
   {
      if(prime[i]!=0)
      { 
         for(j=2;j<n;j++)
         {
            {
               prime[j*prime[i]]=0;
               if(prime[i]*j>n)
                  break;    
            }
         }
      }
   }
   for(i=0;i<n;i++) // Prints the prime numbers
      if(prime[i]!=0)
      {
         printf("%d\n"prime[i]);
      }
      return(0);
   }
}

回答1:

You're allocating a huge array in stack:

int prime[2000000]={};

Four bytes times two million equals eight megabytes, which is often the maximum stack size. Allocating more than that results in segmentation fault.

You should allocate the array in heap, instead:

int *prime;
prime = malloc(2000000 * sizeof(int));
if(!prime) {
    /* not enough memory */
}
/* ... use prime ... */
free(prime);

回答2:

Here is my implementation.

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

int* sieve(int n) {
  int* A = calloc(n, sizeof(int));
  for(int i = 2; i < (int) sqrt(n); i++) {
    if (!A[i]) {
      for (int j = i*i; j < n; j+=i) {
        A[j] = 1;
      }
    }
  }
  return A;
}

I benchmarked it for the first 1,000,000,000 numbers on an i5 Kaby Lake.

🐻 time ./sieve 1000000000
./sieve 1000000000  16.21s user 1.05s system 99% cpu 17.434 total

I simply translated this pseudocode from Wikipedia.

回答3:

Here was my implementation (Java) much simpler in that you really only need one array, just start for loops at 2.

edit: @cheesehead 's solution was probably better, i just read the description of the sieve and thought it would be a good thought exercise.

      // set max;
      int max = 100000000;

      // logic
      boolean[] marked = new boolean[max]; // all start as false
      for (int k = 2; k < max;) {
         for (int g = k * 2; g < max; g += k) {
            marked[g] = true;
         }
         k++;
         while (k < max && marked[k]) {
            k++;
         }
      }

      //print
      for (int k = 2; k < max; k++) {
         if (!marked[k]) {
            System.out.println(k);
         }
      }

来源：https://stackoverflow.com/questions/7921456/sieve-of-eratosthenes

标签

sieve-of-eratosthenes

sieve