“向量 <bool> MSVC中出现“迭代器不可取消引用”错误，但在使用g ++进行编译时效果最佳

Question

I was writing sieve of eratosthenes algorithm in MSVC using a vector of bools(since I intended on making the array/vector dynamic with user input)

My code:

#include<iostream>
#include<cmath>
#include<vector>

void sieve(std::vector<bool>& prime)
{
    long long size = prime.size();
    long long sq = (long long)sqrt(size);
    if (size >= 2)
        prime[0] = prime[1] = false;
    for (long long i = 2; i <= sq; ++i)
        if (prime[i])
            for (long long j = i*i; j <= size; j += i)
                prime[j] = false;
}

int main()
{
    int m, n;
    std::cout << "Enter first number: ";
    std::cin >> m;
    std::cout << "Enter second number: ";
    std::cin >> n;
    std::vector<bool> prime(n, true);
    sieve(prime);
    for (long long i = m; i <= n; ++i)
        if (prime[i])
            std::cout << i << std::endl;
}

I stumbled upon a run time error in MSVC

MSVC Error

But this code works perfectly when compiled using g++. I don't know whats wrong. Any help would be appreciable

Thank you

Answer 1

for (long long i = m; i <= n; ++i) and for (long long j = i*i; j <= size; j += i) will both run past the end of the vector as vector_name[vector_size] is 1 past the end of the elements in the vector. This is undefined behavior and you were unlucky that it worked on g++. Some people never bother to try and compile on another compiler to see if they get the same results and If you hadn't there would have been a silent bug in your "working code".

Change the loops to for (long long i = m; i < n; ++i) and for (long long j = i*i; j < size; j += i) and you will no longer run past the end of the vector.

Answer 2

for (long long j = i*i
  prime[j] = false; 

This is your problem the position is greater than your size of your vector.

Another thing that I have noticed is the size of your vector should be n*n :

std::vector<bool> prime(n*n, true);
void sieve(std::vector<bool>& prime,int m,int n)
{
  long long size = prime.size();
  // long long sq = (long long)sqrt(size); you can use n for this so you don't have to make another variable.
  if (size >= 2)
    prime[0] = prime[1] = false;

  for (long long i = m; i < n; ++i)
  {
    if (prime[i])
    {
      for (long long j = i*i; j < n*n; j += i)
      {
        prime[j] = false;
      }
    }
  }
}
int main()
{
  int m, n;
  std::cout << "Enter first number: ";
  std::cin >> m;
  std::cout << "Enter second number: ";
  std::cin >> n;
  std::vector<bool> prime(n*n, true);
  sieve(prime,m,n);
  for (long long i = m; i <= n; ++i)
  {
    if (prime[i])
    {
      std::cout << i << std::endl;
    }
  }
}

this should work ;), and don`t forget to include the headers.