Thread: processing dynamic arrays

Not sure if you are asking if you can do it without creating a "dynamic" array or not, but in case you are asking if you can do it via a static array you can simply just do:

Code:

#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;

int main(void)
{
	vector<int> ivec;
	int input;

	while (cin >> input) {
			ivec.push_back(input);
	}

	int iarray[ivec.size()];
	copy(ivec.begin(), ivec.end(), iarray);

	// test: print
	for (int i = 0; i < ivec.size(); i++) {
			cout << iarray[i] << endl;
	}

	return 0;
}

Not sure what your asking though

I reread my question after the frivolous adjective has been given to it.

You are right, as was Tonto. It's a confusing question. And more so because I didn't even remember that *(p + i) yields an lvalue. As such, the final loop could have been replaced with...

Code:

    for (vector<int>::size_type i = 0; i != iv.size(); ++i) {
        *(p + i) = iv[i];
        cout << *(p + i) << "\t";
    }

Much simpler...
Dunno, I drew a blank or something. Couldn't see that when I was answering the exercise.