Why doesn't std::queue shrink its memory after popping elements?

Question

I wrote a small program that uses std::queue

queue<double> the_queue;

for(int i = 0; i < 100; i++)
{
    for(int j = 0; j < 5000; j++)
    {
        double d = 1.0 * (rand() % 1000);
        the_queue.push(d);
    }
}

printf("Done pushing\n");

while(!the_queue.empty())
{
    the_queue.pop();
}

printf("Done popping\n");

I put 2 breakpoints at printf("Done pushing\n"); and printf("Done popping\n"); and check program's memory usage (showed in Task Manager) when the breakpoints are hit. At Done pushing, the memory usage is ~ 34 MBs, but at Done popping the memory usage is still ~ 34 MBs. That surprises me!

Why is this? Is there any way to overcome this?

There's also the entire matter that even if you return the memory as free to the C++ (and, by extension, C) runtime, that runtime will probably not give it right back to the OS, so the program will still consume that memory as far as the OS is concerned. — Angew is no longer proud of SO, Jul 05 '16 at 08:42

score 11 · Accepted Answer · edited Nov 12 '20 at 14:43

Basically std::queue is an Adapter Container - it is not a container by its own, but a thin wrapper around other container.

For example, lets take a look at the queue signature:

template <class T, class Container = deque<T> > class queue;

as you can see, T is the type of the element stored in the queue, and Container is the underlying container.

and this is the answer to your question: different containers handles memory differently. the underlying deque may or may not shrink, but it is up to the deque inside to decide.

you can use std::list as your underlying container as well. in this case, each pop deletes the underlying list node memory.

you can also write your own or modify existing container to match your own memory-management patterns. your container needs to support some methods (such as push_back, pop_front) which you can read in the relevant online documentation.

Here is an example to a deque adapter which shrinks in capacity every 1024 pop calls:

template<class T>
class DequeAdapter{
    
private:
    std::deque<T> m_Deque;
    size_t m_PopCount;

public:
    DequeAdapter():
        m_PopCount(0){}
    
    bool empty() const noexcept{
        return m_Deque.empty();
    }
    
    size_t size() const noexcept{
        return m_Deque.size();
    }
    
    T& front() noexcept{
        return m_Deque.front();
    }
    
    const T& front()const noexcept{
        return m_Deque.front();
    }
    
    T& back() noexcept{
        return m_Deque.back();
    }
    
    const T& back()const noexcept{
        return m_Deque.back();
    }
    
    void push_back(const T& t){
        return m_Deque.push_back(t);
    }
    
    void push_back(T&& t){
        return m_Deque.push_back(std::move(t));
    }
    
    void pop_front(){
        m_Deque.pop_front();
        m_PopCount++;
        if (m_PopCount%1024U == 0U){
            m_Deque.shrink_to_fit();
        }
    }

}


template <class T>
using LeanQueue = std::queue<T,DequeAdapter<T>>;

Do note however, that shrinking in capacity means moving or copying the queue elements to the new lean chunk, the memory consumption will be smaller, but the performance may degrade.

score 5 · Answer 2 · answered Jul 05 '16 at 08:43

Any memory the queue manages will be released when the queue goes out of scope.

However, even then memory may not be released back to the OS because the standard library makes the assumption that if you used the memory before, you may well need it again.

The specifics of this are taken care of in malloc/free in the specific c runtime library your program is linked against.

Is this an embedded system where memory is tight? (in which case perhaps consider fixed-size containers), or is it running on a server/desktop/ipad? (in which case tell your boss to stop worrying about things he doesn't understand).

score 1 · Answer 3 · answered Jul 05 '16 at 09:02

Allocating and deallocating memory could become quite an overhead with a container that is regularly used over and over. For this reason containers are allowed to grow to whatever extent they are used. If you want to reclaim the memory you have to do it explicitly. Often that involved allowing the container to go out of scope because the Standard (for some reason) does not give a function to release the memory. There is a function to maybe/maybe not release unused memory on some containers (shrink_to_fit()) but its no guarantee.

There is, thankfully, an idiom that is used to reduce a container to its starting amount of memory called the swap and reset.

You basically create a new empty container and swap() its (empty) contents with your working container.

// create a temporary empty container and swap its
// contents with the working container
std::queue<double>().swap(q);

score 0 · Answer 4 · answered Jul 05 '16 at 08:32

0

Because pop just decreases its size, not is capacity.

Then if you put other elements in the queue, it will not have to allocate memory, it will just use the already-allocated one.

answered Jul 05 '16 at 08:32

Mattia F.

1,720
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I implemented a similar thing in a commercial program and the similar situation has occurred. My boss asked me why the program consumes too much memory. Can I do something? – duong_dajgja Jul 05 '16 at 08:33
Maybe changing STL container? – Mattia F. Jul 05 '16 at 08:36
2

@duong_dajgja perhaps `std::queue().swap(the_queue);` – Piotr Skotnicki Jul 05 '16 at 08:38

Why doesn't std::queue shrink its memory after popping elements?

4 Answers4