Question

I have this example code that doesn’t compile:

#include <iostream>
#include <vector>

using std::endl; using std::cin; using std::cout;
using std::vector;


int main()
{
    vector<int> vi, seg;
    vi.push_back(3);
    vi.push_back(4);
    vi.push_back(5);
    vi.push_back(6);
    vi.push_back(7);
    vector<int>::const_iterator ci_start = vi.begin();
    vector<int>::const_iterator ci_actual = vi.begin();
    while (ci_actual != vi.end())
    {   
        seg = vi(ci_start, ci_actual);
        cout << "......" ;
        for (const vector<int>::const_iterator ci = vi.begin();
                                    ci != vi.end(); ++ci)
            cout << *ci << " ";
        cout << endl;

        ++ci_actual;
    }   
}

What I want is to pass increasing parts of vector vi (that is going to take some million elements) to a function (not shown). These segments are considered from the begining of vi up to where actual is at the moment.
For this, I’ve declared seg that was supposed to have part of vi.

Desired output would be:

3
3 4
3 4 5
3 4 5 6
3 4 5 6 7

Is seg not supposed to hold a copy of elements of vi denoted by iterators ci_start and ci_actual?

To pass the ever increasing part of vi to the function I refered (absent), can I do it faster without copying the elements of vi to another vector and just pass a reference to a segment of vi?

Answer 1

int main()
{
    vector<int> vi;

    vi.push_back(3);
    vi.push_back(4);
    vi.push_back(5);
    vi.push_back(6);
    vi.push_back(7);

    //the end of the sequence 
    vector<int>::iterator seq_end = vi.begin() + 1;

    while (seq_end != vi.end())
    {
        cout << "......" ;
        for (vector<int>::iterator ci = vi.begin(); ci != seq_end; ++ci)
            cout << *ci << " ";
        cout << endl;

        any_function(vi.begin(), seq_end);

        ++seq_end;
    }   
}

void any_function(const vector<int>::iterator& seq_start,
                  const vector<int>::iterator& seq_end)
{
    ...
}

this code make it, analyze and repair your mistakes.