Re: vector.clear() and vector copying

From:

"=?iso-8859-1?q?Erik_Wikstr=F6m?=" <eriwik@student.chalmers.se>

Newsgroups:

comp.lang.c++

Date:

23 Mar 2007 00:20:08 -0700

Message-ID:

<1174634408.248795.97400@n59g2000hsh.googlegroups.com>

On 23 Mar, 01:44, "Jess" <w...@hotmail.com> wrote:

Thanks for all the replies and sorry for not being too specific, here
is my code. :)

#include<iostream>
#include<vector>
#include<string>

using namespace std;

int main(){
  vector<int> v1;
  vector<int> v2;
  vector<string> v3;
  v1.push_back(1);
  v1.push_back(2);
  v2.push_back(3);
  v3.push_back("a");

  cout << v1.size() << endl; //output size is indeed 2
  cout << v2.size() << endl; //output size is indeed 1
  cout << v3.size() << endl; //output size is indeed 1
  v1.clear(); //clear v1
  cout << v1.size() << endl; //v1's size is indeed 0
  cout << v1[0] << endl; //output is 1, garbbage?

  v3.clear();
  cout << v3.size() << endl; //size is indeed 0
  cout << v3[0] << endl; //null string on one machine, and
"Segmentation fault" on another

  //vector copying
  v1.push_back(1); //put back the values again
  v1.push_back(2);

  v1 = v2; //copy v2 to v1
  cout << v1.size() << endl; //v1's size has been decreased to 1
  return 0;

}

If I used vector.at() as suggested by Erik for v1.at(0), then I got
"Abort". So, is vector.at() a much better way to access vector's
content than indices?

Yes, it checks whether the element you are trying to access exist or
not, if it doesn't it will throw an exception. However depending on
your usage of the vector there might be situations where you know for
sure that the element is in the vector, and then the [] operator can
be used, on example of such a situation is when looping over all
members of the vector:

for (int i = 0; i < v.size(); ++i)
std::cout << v[i] << "\n";

--
Erik Wikstr=F6m