Why no see by screen one in c++ if i am using a iterator

Question

I don't know why there is no display to screen with the number one .
The iterator is equal to one, and 1 is > that 0. Please Explain it to me.

int main()
{
   bool hola;
   vector<bool> v;
   vector<bool>:: iterator it = v.begin();
   v.push_back (hola);
   cout << "numero tamano: " << int(v.size()) << endl;
   int i = int(v.size());
   if(i>0)
   }
      cout << *it << " this";
   }

   cout << "fin\n";

   return 0;
}

it only output by screen is this:

numero tamano: 1

Answer 1

You initialize your iterator before your push back. That way your iterator is pointing to somewhere you don't have access Thats why you get a segmentation You will have to initialize the iterator after pushback. It will work fine.

bool hola;
vector<bool> v;
v.push_back (hola);
vector<bool>:: iterator it = v.begin();
cout << "numero tamano: " << int(v.size()) << endl;
int i = int(v.size());
while(it!=v.end())
{
    cout << *it << " this";
    it++;
}

Answer 2

You save an iterator in the line

vector<bool>:: iterator it = v.begin();

Then you perform an operation that (potentially (probably)) invalidates iterators (may not if you had previously called .reserve - but you didn't)

   v.push_back (hola);

Then you use the stored (now potentially (probably) invalid) iterator

  cout << *it << " this";

Don't do that. Accessing an invalidated iterator is Undefined Behaviour and your entire program now has no meaning.