Size in bytes of a structure containing unordered_map

Need to find the exact size in bytes, occupied by a tree data structure that I have implemented. Node Structure is as follows

struct Node
{
    int  word;       
    int   count;       
    unordered_map<int, Node*> map;       

}node;   

What I have been doing is size(int)*2(for word and count) + map.bucket_count() * (sizeof(int) + sizeof(Node*)) and repeatedly do this for each node. Is this the correct way of doing this if i am neglecting the element overhead of storage in the unordered_map?

Also, if I am correct map.bucket_count() gives the number of buckets that are currently allocated that is including the pre- allocated ones. Should I be using map.size() instead which will ignore the pre-allocated buckets?

Or instead of all this, is it better to use tools like MemTrack to find th开发者_C百科e memory used?

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Or instead of all this, is it better to use tools like MemTrack to find the memory used?

Yes. There's no telling from the outside how much memory an unordered_map or other complex, opaque object takes, and a good memory profiler might also show you how much overhead the memory allocator itself takes.