Question

I have a sorted std::vector of relative small size ( from 5 to 20 elements ).
I used std::vector since the data is continuous so I have speed because of cache.
On a specific point I need to remove an element from this vector.

I have now a doubt: which is the fastest way to remove this value between the 2 options below?

1. setting that element to 0 and call sort to reorder: this has complexity but
   elements are on the same cache line.
2. call erase that will copy ( or memcpy who knows?? ) all elements after it
   of 1 place ( I need to investigate the behind scense of erase ).

Do you know which one is faster?

I think that the same approach could be thought about inserting a new element without hitting the max capacity of the vector.

Regards

AFG

Answer 1

If you don’t care about the order of elements, you may be able to swap the element with the last one.

void Remove( std::vector<Object> &vec, iterator i ) {
    iterator last = vec.end()-1;
    if (i != last)
        std::swap( *i, *last );
    vec.erase( last );
}

You mention setting the element to 0. If that means you have pointers, then you may not need the swap:

void Remove( std::vector<Object *> &vec, iterator i ) {
    vec[i] = vec.back();
    vec.erase( vec.end()-1 );
}

If you do care about the order, then your second option of using using erase() will preserve it and do the minimum amount of work. It will almost certainly be faster than resorting.