Re: The most efficient way for fill a vector from an array?

From:
Hakusa <Hakusa@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 9 Mar 2009 05:11:05 CST
Message-ID:
<60d3e49b-f602-41bc-a6f4-77757ea3c89b@z1g2000yqn.googlegroups.com>
On Mar 8, 6:41 am, towi <towit...@gmail.com> wrote:

I wonder, what is the most efficient STL way to move the values from
an array into a vector of the same kind?

My general idea would be that there might be something like
(pseudocode):

   std::vector<int> vec;
   int arr[10] = { 1,2,3, 2,3,4, 3,4,5, 10} ;
   std::move( vec.begin(), vec.end(), arr, arr+10); // vec := arr


Actually, you're over thinking this.

     int arr[] = { 1,2,23,3,34,35,345,34,53 };
     std::vector<int> vec( arr, arr + sizeof(arr)/sizeof(int) );

std::vector's ctor does this for you. And your std::move is covered by
std::copy if I understand your intention.

But, if you're into C++0x, g++ 4.4 can do the fallowing:
     std::vector<int> vec = { 1,2,3,4,5 };

The C++0x technique might be theoretically more efficient, but I don't
think it makes a difference. Compiling the fallowing:

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <iterator>

int main()
{
#ifdef USE_NO_OX
     int x[] = { 1,2,3,4,45,6 };
     std::vector<int> vec( x, x + sizeof(x)/sizeof(int) );
#else
     std::vector<int> vec = { 1,2,3,4,45,6 };
#endif

     std::copy (
         vec.begin(), vec.end(),
         std::ostream_iterator<int>( std::cout, " " )
     );
}

with "g++ main.cpp -std=c++0x [-D USE_NO_OX]; strip a.out". The text
in the brackets was used to define USE_NO_OX on builds where I wanted
to use the previous way. I found that after striping, there was no
difference in the two using any -O level.

I think this is about as efficient as it gets.

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