如何在 C++ 11 及更高版本中将 std::map 和 std::unordered_map 的资源手动释放为 std::vector

Question

Assume we have a vector std::vector<int> v and let's assume that some resources are allocated to it. To my knowledge, v.clear() and v.shrink_to_fit() releases all resources allocated to v . I am wondering if there exist similar operations for std::map and std::unordered_map that release all resources manually. I can only find a member function clear() for these two templates. Can someone explain why there is no shrink_to_fit() for the latter two templates?

Answer 1

There is no shrink_to_fit() in std::map , because it would be useless.

In std::vector , to ensure amortized constant insertion time required by standard, implementation can allocate more memory than is currently necessary for future storage (so that it doesn't have to reallocate everything each push_back() ). Most implementations allocate 2*size() if current capacity() would be exceeded.

shrink_to_fit() asks to release that extra memory to make size() == capacity() (but it's not guaranteed that this will actually happen).

Now, std::map is usually implemented as a red-black tree. Adding an element into such structure is just creating new node and a bit of pointer magic. It will not involve reallocation of other nodes and you cannot speed it up by pre-allocating some memory. shrink_to_fit() doesn't make sense, because there is nothing to shrink.

Update after dewaffled 's comment: For std::unordered_map there's a rehash() method which may decrease size of hash table by recalculating it, but similarly to shrink_to_fit() it's not a guaranteed result.

Answer 2

I'm going to assume you're talking about dynamic maps; the shrink_to_fit() wouldn't make sense because the map is only as big as its linked elements. My understanding is that there isn't 'empty' nodes for a map, like you could have empty fields in a vector.