In C++11 shrink_to_fit
was introduce to complement certain STL containers (e.g., std::vector
, std::deque
, std::string
).
Synopsizing, its main functionality is to request the container that is associated to, to reduce its capacity to fit its size. However, this request is non-binding, and the container implementation is free to optimize otherwise and leave the vector with a capacity greater than its size.
Furthermore, in a previous SO question the OP was discouraged from using shrink_to_fit
to reduce the capacity of his std::vector
to its size. The reasons not to do so are quoted below:
shrink_to_fit
does nothing or it gives you cache locality issues and it's O(n) to execute (since you have to copy each item to their new, smaller home). Usually it's cheaper to leave the slack in memory. @Massa
Could someone be so kind as to address the following questions:
std::vector
).shrink_to_fit
after-all? C++ std::vector Reducing the Capacity of a Vector In C++11 we can use the shrink_to_fit() member function for a similar effect: v. shrink_to_fit();
No. pop_back() will not shrink the capacity of vector.
We can set the size of a Vector using setSize() method of Vector class. If new size is greater than the current size then all the elements after current size index have null values. If new size is less than current size then the elements after current size index have been deleted from the Vector.
So, in this way, we can use the vector::shrink_to_fit() function to shrink the capacity of the vector to its size.
Do the arguments in the quotation hold?
Measure and you will know. Are you constrained in memory? Can you figure out the correct size up front? It will be more efficient to reserve
than it will be to shrink after the fact. In general I am inclined to agree on the premise that most uses are probably fine with the slack.
If yes, what's the proper way of shrinking an STL container's capacity to its size (at least for std::vector).
The comment does not only apply to shrink_to_fit
, but to any other way of shrinking. Given that you cannot realloc
in place, it involves acquiring a different chunk of memory and copying over there regardless of what mechanism you use for shrinking.
And if there's a better way to shrink a container, what's the reason for the existence of shrink_to_fit after-all?
The request is non-binding, but the alternatives don't have better guarantees. The question is whether shrinking makes sense: if it does, then it makes sense to provide a shrink_to_fit
operation that can take advantage of the fact that the objects are being moved to a new location. I.e., if the type T
has a noexcept(true)
move constructor, it will allocate the new memory and move the elements.
While you can achieve the same externally, this interface simplifies the operation. The equivalent to shrink_to_fit
in C++03 would have been:
std::vector<T>(current).swap(current);
But the problem with this approach is that when the copy is done to the temporary it does not know that current
is going to be replaced, there is nothing that tells the library that it can move the held objects. Note that using std::move(current)
would not achieve the desired effect as it would move the whole buffer, maintaining the same capacity()
.
Implementing this externally would be a bit more cumbersome:
{ std::vector<T> copy; if (noexcept(T(std::move(declval<T>())))) { copy.assign(std::make_move_iterator(current.begin()), std::make_move_iterator(current.end())); } else { copy.assign(current.begin(), current.end()); } copy.swap(current); }
Assuming that I got the if condition right... which is probably not what you want to write every time that you want this operation.
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