Iterating over vectors - speed difference

From:
Arijit <pal_ari@yahoo.co.in>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 31 Dec 2009 13:06:56 CST
Message-ID:
<12929b26-62ce-48fe-ae44-d94f8f50d2e6@z41g2000yqz.googlegroups.com>
Hi

I am comparing the runtime of iterating through a vector using 3
different approaches -
1) Indexing -> i = 0; i < vec.size();
2) Iterator -> i = vec.begin(), i != vec.end()
3) Summing using Accumulate

Here are the runtimes on my computer. I am compiling using VC++ 2008,
Release mode build. Runtimes are pretty stable from run to run.

1) Indexing -> 1.884 s
2) Iterator -> 8.725 s
3) Accumulate -> 1.8 s

As expected, accumulate is the fastest of them all, but only by a
narrow margin. The shocker is using iterators is nearly 5 times slower
than indexing, whereas I expected the two to be nearly at par. I don't
understand why this should happen.

Is this a case specific to VC++ / Dinkumware libraries ? Or is this
expected based on the C++ standard ? Or have I made an error in my
benchmarking ?

Thanks for your responses!

-Arijit

#include <iostream>
#include <vector>
#include <numeric>
#include <ctime>

using namespace std;

int main()
{
    const int maxnum = 1000000;
    const int iter = 1000;

    vector<int> num;
    num.reserve(maxnum);

    for(int i = 0; i < maxnum; ++i)
        num.push_back(i);

    clock_t start = clock();

    long long sum = 0;
    for(int i = 0; i < iter; ++i)
        for(unsigned int j = 0; j < num.size(); ++j)
            sum += num[j];

    cout << sum << endl;
    clock_t mid = clock();

    sum = 0;
    for(int i = 0; i < iter; ++i)
        for(vector<int>::iterator j = num.begin(); j != num.end(); ++j)
            sum += *j;

    cout << sum << endl;
    clock_t late = clock();

    sum = 0;
    for(int i = 0; i < iter; ++i)
        sum += accumulate(num.begin(), num.end(), 0ll);

    cout << sum << endl;
    clock_t end = clock();

    cout << endl << static_cast<double>(mid - start) / CLOCKS_PER_SEC;
    cout << endl << static_cast<double>(late - mid) / CLOCKS_PER_SEC;
    cout << endl << static_cast<double>(end - late) / CLOCKS_PER_SEC <<
endl << endl;

    return 0;
}

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