Square Root Code C++ without sqrt() - c++

Question

I have to create a code where the user inputs a number which is a perfect square, and I have to show its root. I've made this code, but I'm getting Segmentation Fault 11 , in this piece: int j = squareRootVector[i];
squareRoot.push_back(j);.
I can't change the code too much, so is there a way that I can do that?
#include <iostream>
#include <vector>
using namespace std;
int main() {
cout <<
"Enter the number:\n";
int input;
int number = input;
int divider = 2;
vector<int> squareRootVector;
vector<int> squareRoot;
cin >> number;
for(int divider = 2; number > 1; divider++) {
while((number % divider) == 0) {
number /= divider;
cout << number << endl;
squareRootVector.push_back(divider);
}
}
for(int i = 0; i < squareRootVector.size(); i++) {
cout << squareRootVector[i] << " ";
/*******PROBLEM*******/
if(squareRootVector[i] == squareRootVector[i+1]) {
int j = squareRootVector[i];
squareRoot.push_back(j);
}
/*********************/
}
int root;
for (int i = 0; squareRoot.size(); i++) {
root = root * squareRoot[i];
}
cout << "Square Root of " << input << " is: " << root << endl;
return 0;
}

Answer 1

The behaviour on accessing squareRootVector[i+1] with i just one below size (which your loop constaint allows) is undefined.
Consider writing
for (std::size_t i = 1; i < squareRootVector.size(); i++) {
instead, and rebasing the for loop body accordingly. I've also slipped in a change of type for i.

Answer 2

Shortly, the problem is that the last cycle in the last "for":
for(int i = 0; i < squareRootVector.size(); i++)
has the following line in it:
squareRootVector[i] == squareRootVector[i+1];
This is an "out of limits" error: squareRootVector only has squareRootVector.size() elements (let's say n), and the elements are indexed from 0 to n-1.
squareRootVector[i+1] in the last cycle points one element after the last one of squareRootVector, which is undefined behavior.

Answer 3

Using vector::iterator is proper way.
for(vector<int>::iterator it = squareRootVector.begin(); it != squareRootVector.end(); ++it)
{
if( (it+1) == squareRootVector.end() )
{
//what to do if there's no next member???
break;
}
if( *it == *(it+1) )
{
squareRoot.push_back(*it);
}
}

Answer 4

Thanks for the answers, guys. I've ended up with this code:
#include <iostream>
#include <vector>
using namespace std;
int main() {
cout << "Enter the number:\n";
int input = 0;
int number = 0;
cin >> input;
number = input;
int divider = 2;
vector<int> squareRootVector;
vector<int> squareRoot;
for(int divider = 2; number > 1; divider++) {
while((number % divider) == 0) {
number /= divider;
squareRootVector.push_back(divider);
}
}
int vectorSize = squareRootVector.size() - 1;
for(int i = 0; i < vectorSize; i++) {
if(squareRootVector[i] == squareRootVector[i+1]) {
int j = squareRootVector[i];
squareRoot.push_back(j);
}
}
int root = 1;
for (int i = 0; i < squareRoot.size(); i++) {
root = root * squareRoot[i];
}
cout << "Square Root of " << input << " is " << root << endl;
return 0;
}