Benefits of using reserve() in a vector - C++

user_812849 0
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What is the benefit of using reserve when dealing with vectors. When should I use them? Couldn't find a clear cut answer on this but I assume it is faster when you reserve in advance before using them.

What say you people smarter than I?

vector c++ performance
8 Answers

This excellent article deeply explains differences between deque and vector containers. Section "Experiment 2" shows the benefits of vector::reserve().

user_16369070

Although its an old question, Here is my implementation for the differences.

#include <iostream>
#include <chrono>
#include <vector>

using namespace std;

int main(){
    vector<int> v1;
    chrono::steady_clock::time_point t1 = chrono::steady_clock::now();
    for(int i = 0; i < 1000000; ++i){
        v1.push_back(1);
    }
    chrono::steady_clock::time_point t2 = chrono::steady_clock::now();
    chrono::duration<double> time_first = chrono::duration_cast<chrono::duration<double>>(t2-t1);
    cout << "Time for 1000000 insertion without reserve: " << time_first.count() * 1000 << " miliseconds." << endl;

    vector<int> v2;
    v2.reserve(1000000);
    chrono::steady_clock::time_point t3 = chrono::steady_clock::now();
    for(int i = 0; i < 1000000; ++i){
        v2.push_back(1);
    }
    chrono::steady_clock::time_point t4 = chrono::steady_clock::now();
    chrono::duration<double> time_second = chrono::duration_cast<chrono::duration<double>>(t4-t3);
    cout << "Time for 1000000 insertion with reserve: " << time_second.count() * 1000 << " miliseconds." << endl;
    return 0;
}

When you compile and run this program, it outputs:

Time for 1000000 insertion without reserve: 24.5573 miliseconds.

Time for 1000000 insertion with reserve: 17.1771 miliseconds.

Seems to be some improvement with reserve, but not that too much improvement. I think it will be more improvement for complex objects, I am not sure. Any suggestions, changes and comments are welcome.

user_21037320

It's always interesting to know the final total needed space before to request any space from the system, so you just require space once. In other cases the system may have to move you in a larger free zone (it's optimized but not always a free operation because a whole data copy is required). Even the compiler will try to help you, but the best is to to tell what you know (to reserve the total space required by your process). That's what i think. Greetings.

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