C++求Fib数列

int fib(int pos)  {    int elem = 1;    int n1 = 1, n2 = 1;    for (int i = 3; i <= pos; i++)    {      elem = n2 + n1;      n1 = n2;      n2 = elem;    }    return elem;  }

bool fib(int pos, int &elem)  {    if(pos < 0 || pos > 1024)    {      elem = 0;      return false;    }    elem = 1; //注意：定义只能有1次    int n1 = 1, n2 = 1;    for (int i = 3; i <= pos; i++)    {      elem = n2 + n1;      n1 = n2;      n2 = elem;    }    return true;  }

int main()  {        int pos;    cout <<"Please enter a position: ";    cin >> pos;      int elem;    if(fib(pos, elem))    {      cout << "element # " << pos         << " is " << elem << endl;    }    else      cout << "Sorry. Couldn't calculate element #"         << pos <<endl;  }

const vector<int>* fib_new(int size)  {    const int max_size = 1024;    static vector<int> elems;      if(size <= 0 || size >= max_size)    {      cerr << "fib_new(): oops:invalid size:"         << size << "-- can't fulfill request./n";      return 0;    }    for(int ix = elems.size(); ix < size; ix++)    {      if (ix == 0 || ix == 1)        elems.push_back(1);      else        elems.push_back(elems[ix - 1] + elems[ix - 2]);    }    return &elems;  }

    const vector<int> *result=fib_new(5);    cout << result->back();        const vector<int> *result=fib_new(5);    cout << result->at(4)<< endl;    //这个应该怎么多次调用返回，这个还没明白，留个记号。