What happens under the hood of vector::push_back memory wise?
My question is regarding the effect of vector::push_back, I know it adds an element in the end of the vector but what happens underneath the hood?
IIRC memory objects are allocated in a sequential manner, so my question is whether vector::push_back simply allocates more memory immediately after the vector, and if so what happens if there is not enough free memory in that location? Or perhaps a pointer is added in the "end" to cause the vector to "hop" to the location it continues? Or is it simply reallocated through copying it to another location that has enough space and the old copy gets discarded? Or maybe something else?
Answer
If there is enough space already allocated, the object is copy constructed from the argument in place. When there is not enough memory, the vector will grow it's internal databuffer following some kind of geometric progression (each time the new size will be k*old_size with k > 1[1]) and all objects present in the original buffer will then be moved to the new buffer. After the operation completes the old buffer will be released to the system.
In the previous sentence move is not used in the technical move-constructor/ move-assignment sense, they could be moved or copied or any equivalent operation.
[1] Growing by a factor k > 1 ensures that the amortized cost of push_back is constant. The actual constant varies from one implementation to another (Dinkumware uses 1.5, gcc uses 2). The amortized cost means that even if every so often one push_back will be highly expensive (O(N) on the size of the vector at the time), those cases happen rarely enough that the cost of all operations over the whole set of insertions is linear in the number of insertions, and thus each insertion averages a constant cost)