Sieve of Eratosthenes c++ implementation error - c++

Question

I wrote this implementation of the Sieve of Eratosthenes in c++, but whenever the code reaches 59 (the 16th prime), it stops working. It can only reach 37 on my older machine. When I debug the program, all the variables seem to be working fine; the program simply crashes. Here it is: (I know it has a lot of comments, and lots are unnecessary.)
// Problem 7:What is the 10 001st prime number?
/*This program generates a list of prime numbers and uses the Sieve of Eratosthenes to find them.
*This is done by crossing out multiples of the first known prime (2), taking the first number
*not yet crossed out, and setting that as the next prime to "sieve" with.
*/
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int placeholder; //added at the end to prevent window from closing
const int sieve_size = 10000; //amount of #s to sieve through
const int solution_number = 10001; //# of primes to generate
long long solution; //the 10 001st prime #
long long current_sieve; //current # sieving with
int number_of_primes = 1; //we know the first prime -- 2
long long *primes = new long long [number_of_primes];
primes[0] = 2; //2 is the first prime
bool flag_havePrime = 0; //whether we have our next prime yet; used when saving a prime
bool sieve[sieve_size] = {0}; //0 is "could-be-prime" (not composite), 1 is composite.
sieve[0] = 1; //0 and 1 are not prime #s
sieve[1] = 1;
for (int i = 0; number_of_primes <= solution_number; i++) //each loop sieves with a different prime
{
current_sieve = primes[i];
//This next loop sieves through the array starting with the square of the number we will sieve with,
//this optimizes the run time of the program.
for (long long j=current_sieve*current_sieve; j <= sieve_size; j++)
if (j%current_sieve == 0) sieve[j] = 1;
/*This loop gets our next prime by looking through the array until
*it encounters a number not crossed out yet. If it encounters a prime,
*it increments the number of primes, then saves the new prime into
*primes[]. It also prints that prime (for debugging purposes).
*The "for" loop ends when it finds a prime, which it knows by encountering
*the "havePrime" flag. This needs to be reset before every run.
*/
for (long long j = primes[number_of_primes-1]+1; flag_havePrime == 0; j++)
{
if (sieve[j] == 0)
{
number_of_primes++;
primes[number_of_primes-1] = j; //because array counting starts # 0
cout << primes[number_of_primes-1] << endl; //because array counting starts # 0
flag_havePrime = 1;
}
}
flag_havePrime = 0; //resetting this flag
}
solution = primes[number_of_primes-1]; //because array counting starts # 0
delete[] primes;
primes = 0;
cout << "The 10 001st prime number is:\n";
cout << solution << endl;
cin >> placeholder;
return 0;
}
I'm thinking it could be an overflow problem?
Here's an updated snippet with only the changes:
const int sieve_size = 500000;
long long *primes = new long long [solution_number];
Debugging returns a (gasp) heap overflow though, but running the compiled version doesn't. The compiled version stops at 104759, going over by 1. That's probably easy enough to fix. But the program doesn't print the last bit, where it gives you the solution. Weird.

Answer 1

int number_of_primes = 1; //we know the first prime -- 2
long long *primes = new long long [number_of_primes];
This will create a one-element array. I'm pretty certain you'll need something bigger than that for storing the primes.
Specifically, as soon as you start setting values like primes[11] (for example), you're into the realm of undefined behaviour.
Perhaps there's a different variable you may want to consider using for the size in the new statement, nudge, nudge, wink, wink, a nod's as good as a wink to a blind horse :-)
There's a few other problems you also have in that code. The main one is that your sieve itself is only 10,000 elements long. The idea of a sieve is that you take a large number of things and filter out those not matching. For what it's worth, the 10,001st is a touch under 105,000 so your sieve should be at least that big.
Secondly, I've seen people use the square of a number to optimise finding of factors but not in this way:
for (long long j=current_sieve*current_sieve; j <= sieve_size; j++)
if (j%current_sieve == 0) sieve[j] = 1;
What you want is to start at double the current sieve and add it each time, something like:
for (long long j = current_sieve * 2; j < sieve_size; j += current_sieve)
sieve[j] = 1;

Answer 2

Here is a fixed version to compare with:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
const int sieve_size = 1 << 20;
const int solution_number = 10001;
int number_of_primes = 0;
vector<bool> sieve(sieve_size, true);
for (int i = 2; i < sieve_size; i++)
{
if (!sieve[i])
continue;
if (++number_of_primes == solution_number)
{
cout << "The 10 001st prime number is:" << i << endl;
return 0;
}
for (long long j = i*2; j < sieve_size; j += i)
sieve[j] = false;
}
}
Output:
The 10 001st prime number is:104743