C++ STL：为什么分配器不会增加容器的内存占用？

Question

The following code snippet( see on godbolt ) shows that big allocators won't increase the memory footprint of STL containers, but big comparators will. Why is it the case?

// compiled with x86-64 gcc 10.3, -std=c++17
#include <functional>
#include <iostream>
#include <memory>
#include <set>

struct MyLess : public std::less<int>
{
    char dummy[1024];
};

struct MyAllocator : public std::allocator<int>
{
    char dummy[1024];
};

int main()
{
    std::cout << sizeof(std::set<int, MyLess>) << std::endl;  // prints 1064
    std::cout << sizeof(std::set<int, std::less<int>, MyAllocator>) << std::endl;  // prints 48
    return 0;
}

Answer 1

Your allocator is not being used.

By default, std::set receives std::allocator<int> , but it needs to allocate some kind of nodes, not int s. It usesstd::allocator_traits::rebind to get a different allocator for its internal node type.

Pre-C++20 std::allocator has a rebind member type, which you inherit, and which std::allocator_traits::rebind finds. That rebind points to std::allocator , so that's what you get.

Starting from C++20, there's no rebind in std::allocator , so std::allocator_traits::rebind falls back to directly modifying the first template parameter of your allocator, and since it's not a template, you get a compilation error.

A possible solution is to make your allocator a template, and to provide your own rebind (which can be malformed, then the template parameter will be replaced automatically):

template <typename T>
struct MyAllocator : public std::allocator<T>
{
    char dummy[1024];
    struct rebind {}; // Malformed `rebind` to hide the inherited one, if any.
};

Then 1072 is printed for me.