Re: exception when using vector

From:
Alberto Ganesh Barbati <AlbertoBarbati@libero.it>
Newsgroups:
comp.lang.c++.moderated
Date:
4 Dec 2006 15:38:13 -0500
Message-ID:
<LfZch.66879$uv5.564591@twister1.libero.it>
redf0x ha scritto:

Hello,everyone:
      I have contacted a really unreasonable vector exception when I
used its method erase,is there anything wrong int the following code?

               /*sou and dest are both int vector*/
                for(vector<int>::iterator bpiter = sou.begin(); bpiter
!= sou.end();
)
                {
                        if((*bpiter) == 5)
                                (*bpiter) = 3;
                        des.push_back(*bpiter);
                        sou.erase(bpiter);
                }


once you do "sou.erase(bpiter)", bpiter becomes an invalid iterator that
can't be used anymore. In particular, you can't compare it with
sou.end(). Why don't you write your code like this:

  for(vector<int>::iterator bpiter = sou.begin();
    bpiter != sou.end(); ++bpiter)
  {
    if((*bpiter) == 5)
      (*bpiter) = 3;
    des.push_back(*bpiter);
  }
  sou.clear();

In addition to avoid invalidating the iterator, it's much better to
erase all elements in one single call to clear(), rather than one at a
time with erase(). Remember that each time you erase one element from
the front of the vector you have to move *every other element* down one
position.

In this particular case, you can also avoid the loop entirely, as
there's an STL algorithm that performs exactly what you are trying to do:

  std::replace_copy(
    sou.begin(), sou.end(), std::back_inserter(des), 5, 3);
  sou.clear();

In order to increase performances by reducing the need for repeated
allocations, you can add a call to reserve(), as you know that the final
number of elements in des will be des.size() + sou.size().

HTH,

Ganesh

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