Difference between std::remove and erase for vector?

Question

I have a doubt that I would like to clarify in my head. I am aware of the different behavior for std::vector between erase and std::remove where the first physically removes an element from the vector, reducing size, and the other just moves an element leaving the capacity the same.

Is this just for efficiency reasons? By using erase, all elements in a std::vector will be shifted by 1, causing a large amount of copies; std::remove does just a 'logical' delete and leaves the vector unchanged by moving things around. If the objects are heavy, that difference might matter, right?

Answer 1

Is this just for efficiency reason? By using erase all elements in a std::vector will be shifted by 1 causing a large amount of copies; std::remove does just a 'logical' delete and leaves the vector unchanged by moving things around. If the objects are heavy that difference mihgt matter, right?

The reason for using this idiom is exactly that. There is a benefit in performance, but not in the case of a single erasure. Where it does matter is if you need to remove multiple elements from the vector. In this case, the std::remove will copy each not removed element only once to its final location, while the vector::erase approach would move all of the elements from the position to the end multiple times. Consider:

std::vector<int> v{ 1, 2, 3, 4, 5 }; // remove all elements < 5 

If you went over the vector removing elements one by one, you would remove the 1, causing copies of the remainder elements that get shifted (4). Then you would remove 2 and shift all remainding elements by one (3)... if you see the pattern this is a O(N^2) algorithm.

In the case of std::remove the algorithm maintains a read and write heads, and iterates over the container. For the first 4 elements the read head will be moved and the element tested, but no element is copied. Only for the fifth element the object would be copied from the last to the first position, and the algorithm will complete with a single copy and returning an iterator to the second position. This is a O(N) algorithm. The later std::vector::erase with the range will cause destruction of all the remainder elements and resizing the container.

As others have mentioned, in the standard library algorithms are applied to iterators, and lack knowledge of the sequence being iterated. This design is more flexible than other approaches on which algorithms are aware of the containers in that a single implementation of the algorithm can be used with any sequence that complies with the iterator requirements. Consider for example, std::remove_copy_if, it can be used even without containers, by using iterators that generate/accept sequences:

std::remove_copy_if(std::istream_iterator<int>(std::cin),                     std::istream_iterator<int>(),                     std::ostream_iterator<int>(std::cout, " "),                     [](int x) { return !(x%2); } // is even                     ); 

That single line of code will filter out all even numbers from standard input and dump that to standard output, without requiring the loading of all numbers into memory in a container. This is the advantage of the split, the disadvantage is that the algorithms cannot modify the container itself, only the values referred to by the iterators.

Answer 2

std::remove is an algorithm from the STL which is quite container agnostic. It requires some concept, true, but it has been designed to also work with C arrays, which are static in sizes.