CODEWITHSUNDEEP
c++
antho
antho

Reputation: 638

Allocation of a struct vector containing an array

Imagine a structure

struct A {
    int i;
    int arr[1000];
};

std::vector<A> vec;

I cannot find an answer to know if the array "arr" in my structure "A" will take up space on the stack or not at all. I these although the vector will allocate an array of "A" which will be on the heap and will not impact my stack but will my "arr" array impact my stack?

std::vector <A> vec1 (1000000);

std::vector<std::unique_ptr<A>> vec2 (1000000);

Will vec1 "destroy" my stack and better to use vec2?

Edit for the comment.

Why can vec1 allocate more elements than vec2?

#include <iostream>
#include <vector>
#include <memory>

struct A {
    int i;
    int arr[1000];
};

int main() {
  std::vector<std::unique_ptr<A>> vec1;
  std::vector<A> vec2;
  std::cout << vec1.max_size() << '\n' << vec2.max_size() << '\n';
}

Upvotes: 1

Views: 376

Answers (1)

Swift - Friday Pie
Swift - Friday Pie

Reputation: 14688

std::vector is one of AllocatorAwareContainers and default allocator use dynamic allocation (often called heap allocation, which is true for systems with heap-like memory model).

When using those two

std::vector<std::unique_ptr<A>> vec1;
std::vector<A> vec2;

both have own advantages and disadvantages. The vec1 offers arguably better time of resizing container itself because unique_ptr is quite small. vec2 offers instances of A with adjacent storage but resize time would be higher because of move operation used on each element. vec1 allocates only storage for pointers and not for the object itselves, by reaching that number you likely won't have memory left to create instances of A.

For lesser automatic storage (stack) footprint when using vec2 you might want to avoid creation of A temporary at all and use emplace_XXX methods to create instances of A right in vector's storage.

Upvotes: 3

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