C++ Vector memory allocation & run-time?

Question

I've just written a tiny C++ program just to understand how vectors work with memory and what's happen during run-time.

There is my code :

#include <iostream>
#include <cstdio>
#include <ctime>
#include <vector>

int main(){

    clock_t start, end;

    int x;

    std::vector<int> a(5, 100);

    start = clock();

    for(int i = 0 ; i <= 900000000 ; i++){
        x = a[0];
        x = a[1];
        x = a[2];
        x = a[3];
        x = a[4];
    }

    end = clock();

    clock_t duration = end - start;

    double durationPerSec = duration / (double)CLOCKS_PER_SEC;

    std::cout << "Run-time : " << durationPerSec << std::endl;

    return 0;
}

And i got this output :

Run-time : 18.7843

When i write the same code by replacing the vector by a dynamic array the run-time duration is more acceptable :

Run-time : 2.9526

I know this code is quite stupid but i wonder why run-time is so long when i use the vector ? Is that because i use it in the wrong way or just because there's something that i don't understand ?

Thanks for replying.

possible duplicate of [std::vector is so much slower than plain arrays?](http://stackoverflow.com/questions/3664272/stdvector-is-so-much-slower-than-plain-arrays) — , Sep 27 '13 at 14:45
Dynamic array like this one : int *dynamicArray = new int[5]; — FlorentinTh, Sep 27 '13 at 14:47
I suspect, if you actually compile with optimizations on, that your entire loop will disappear, since it can be proven to be "dead code" (it has no effects that anything outside the loop depends on). — twalberg, Sep 27 '13 at 16:16

CS Pei · Accepted Answer · 2013-09-27T14:56:37.153

1

I run it with g++ -O0 a.cc and get

    Run-time : 18.99

But if I use g++ -O2 a.cc

    Run-time : 0

to be more actuate , I run the second with time ./a.out

time ./a.out
Run-time : 0

real    0m0.009s
user    0m0.002s
sys     0m0.002s

I changed the loop to

for(int i = 0 ; i <= 900000000 ; i++){
    a[0] = i ;
    a[1] = i + a[0];
    a[2] =  a[1] + a[2];
    a[3] = i + a[1] + a[2];
    a[4] = i + a[1] + a[2] + a[3];
    x = a[0];
    x = a[1];
    x = a[2];
    x = a[3];
    x = a[4];
}

Then the result of g++ -O2 is

time ./a.out
Run-time : 1.81

real    0m1.817s
user    0m1.811s
sys     0m0.001s

edited Sep 27 '13 at 14:56

answered Sep 27 '13 at 14:44

CS Pei

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yeah but could it be that the whole loop is optimized away now? what would it give a[i]=i; inside the loop or something similar? – Julien Lopez Sep 27 '13 at 14:48
What is the difference between g++ -02 and g++ - O0 ? – FlorentinTh Sep 27 '13 at 15:10
-O0 no optimization. -O2 is usually people use. – CS Pei Sep 27 '13 at 15:56

score 0 · Answer 2 · answered Sep 27 '13 at 14:42

You should measure with optimizations enabled.

operator[] is a member function. When you access elements using [0], it actually goes through a function, so there are more instructions to execute, despite identical notation. In debug, there will be measurable overhead. In release, it is usually unmeasurable.

C++ Vector memory allocation & run-time?

2 Answers2