CODEWITHSUNDEEP
c++stlinsertimplementationcontainers
Avinash
Avinash

Reputation: 13277

STL Container insert element and memory perspective

This question is extension of the question I want to understand how the elements are inserted into the STL Container.

Suppose I have A object;, which I want to insert into any of the STL Container, I understand that there is concept of allocators which handles the memory. But I fail to understand that how the actual object is copied into STL memory. So my object is stored on the stack when I call Container.insert how does STL create copy of this object and stored this objects into its memory.

Any equivalent C++ code would be helpful which simulates the same.

Upvotes: 0

Views: 425

Answers (4)

David Rodríguez - dribeas
David Rodríguez - dribeas

Reputation: 208476

The approach is not that complicated. Basically the container will obtain memory from the allocator and then perform copy-construction (with placement-new over that memory). The easier container to see is vector:

void push_back( T const & value ) {
   ensure_enough_capacity();
   new (end_ptr++) T( value );
}

Where ensure_enough_capacity() determines whether the vector has to grow and does it, that is, it will end up calling the allocator if size()==capacity() when push_back is called.

The next level of complexity is a list, where each node is allocated on its own, and there is some extra information that the library has to manage. In that case the code would look similar to:

void push_back( T const& value ) {
    node* n = allocator::allocate( sizeof(node) );
    new (n) node( value, x, y );
}

Where x and y are the appropriate pointers to initialize the node's next and last pointers (usually would be a pointer to the last node for last and a pointer to a sentry node --invalid beyond the end-- for next), and assuming that this particular constructor will copy-construct the value and then fix all referred pointers.

Ordered associative containers have the extra level of complexity of managing the balanced tree, but the approach is the same: allocate a block big enough to hold the value and the extra information, and then use placement-new to build the node. The rest are details of the data structure.

Upvotes: 3

sehe
sehe

Reputation: 394054

what really happens in terms of memory allocation is this: _it uses the allocator passed in as the allocator template parameter.

 map<int , long , less<int> , myAllocator<pair<int, long> > > myMap;

whatever you make myAllocator do it's allocations from (if any) will be used. This technically means that you could preallocate all the pairs (perhaps from a vector). It also implicates that placement new is being used, just like in the vector case, only that, instead of a single contiguous allocation, your allocation will be called many times for small allocations.

This only leads to the situation where the container cannot guarantee that storage is contiguous (like in the vector case), however, it miht still happen to be contiguous due to the implementation of your allocator

Writing allocators is an advanced topic and it has been addressed in

Upvotes: 0

Emilio Garavaglia
Emilio Garavaglia

Reputation: 20769

The most of the insert functions are prototyped as

void insert(const A& a);

Taking a cost& is -essentially- a way to pass a value without copy it in to the formal parameter.

Container -depending on how they work- have an internal structure that contains As. For example, a list has a 

struct node
{
   A val;
   node* next;
   node* prev;

   node(const A& a) :val(a), next(), perv()
   { }  
};

Insert, at that point does noting more than

void insert(const A& a)
{
   node* n = new(allocator.allocate()) node(a);
   /*set next and prev accordingly to list functionality*/
}

The reference is passed through the node constructor and given to the initializer for the node's embedded A, that's its own copy constructor.

Note that allocator.allocate() atually has to allocate space for node, not just A. For that reason the allocator instance inside the list will be of type typename Allocator::rebind<node>::other, where Allocator is the second template parameter of list, that defaults to std::allocator<A>.

Upvotes: 0

thiton
thiton

Reputation: 36059

The object is copied using the appropriate copy constructor (or, maybe, the move constructor in C++11). Assume that you have pre-allocated an array of N Foo objects and have length objects already in it, the code in std::vector might look like:

void std::vector<Foo>::push_back( const Foo& n ) { 
    new( my_memory+length ) Foo(n);
    ++length;
}

The brackets after the new indicate "placement new", that is, calling new on pre-allocated storage.

Upvotes: 0

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