Why Vector's size() and capacity() is different after push_back()

Question

I just starting to learning vectors and little confused about size() and capacity() I know little about both of them. But why in this program both

Answer 1

The Standard mandates that std::vector::push_back() has amortized O(1) complexity. This means that the expansion has to be geometrically, say doubling the amount of storage each time it has been filled.

Simple example: sequentially push_back 32 ints into a std::vector. You will store all of them once, and also do 31 copies if you double the capacity each time it runs out. Why 31? Before storing the 2nd element, you copy the 1st; before storing the 3rd, you copy elements 1-2, before storing the 5th, you copy 1-4, etc. So you copy 1 + 2 + 4 + 8 + 16 = 31 times, with 32 stores.

Doing the formal analysis shows that you get O(N) stores and copies for N elements. This means amortized O(1) complexity per push_back (often only a store without a copy, sometimes a store and a sequence of copies).

Because of this expansion strategy, you will have size() < capacity() most of the time. Lookup shrink_to_fit and reserve to learn how to control a vector's capacity in a more fine-grained manner.

Note: with geometrical growth rate, any factor larger than 1 will do, and there have been some studies claiming that 1.5 gives better performance because of less wasted memory (because at some point the reallocated memory can overwrite the old memory).