Re: Avoiding copy construction when growing a vector

From:

SG <s.gesemann@gmail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Mon, 18 Oct 2010 19:47:22 CST

Message-ID:

<8ab5904a-0fba-4fcb-9ffc-2199a756d28d@w19g2000yqb.googlegroups.com>

On 18 Okt., 07:41, Marc wrote:

In order to grow such a vector (after checking that capacity() is not
sufficient), you could create a new vector of the new size filled with
default constructed elements, iteratively swap the elements with those
of the old vector, and swap the 2 vectors.

Sounds like a good idea. But "new size" should still be the old size
while the new vector's capacity is at least 1.5 or 2.0 times the old
capacity:

   template<class T>
   void reserve_via_swap(std::vector<T> & vec, size_t cap)
   {
     if (cap<=vec.capacity()) return;
     size_t newcap = std::max<size_t>(cap,vec.capacity()/2*3);
     std::vector<T> tmp;
     tmp.reserve(newcap);
     tmp.resize(vec.size());
     using std::swap;
     for (size_t i=0; i<vec.size(); ++i) {
       swap(vec[i],tmp[i]);
     }
     vec.swap(tmp);
   }

Cheers!
SG

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