why not implement c++ std::vector::pop_front() by shifting the pointer to vector[0]?

Question

Why can't pop_front() be implemented for C++ vectors by simply shifting the pointer contained in the vector's name one spot over? So in a vector containing an array foo, foo is a pointer to foo[0], so pop_front() would make the pointer foo = foo[1] and the bracket operator would just do the normal pointer math. Is this something to do with how C++ keeps track of the memory you're using for what when it allocates space for an array?

This is similar to other questions I've seen about why std::vector doesn't have a pop_front() function, I will admit, but i haven't anyone asking about why you can't shift the pointer.

Answer 1

I started typing out an elaborate answer explaining how the memory is allocated and freed but after typing it all out I realized that memory issues alone don't justify why pop_front isn't there as other answers here suggested.

Having pop_front in a vector where the extra cost is another pointer is justifiable in most circumstances. The problem, in my opinion, is push_front. If the container has pop_front then it should also have push_front otherwise the container is not being consistent. push_front is definitely expensive for a vector container (unless you match your pushes with your pops which is not a good design). Without push_front the vector is really wasting memory if one does lots of pop_front operations with no push_front functionality.

Now the need for pop_front and push_front is there for a container that is similar to a vector (constant time random access) which is why there is deque.