Is there a standard library equivalent of memmove?

Question

The standard library offers std::copy, which can be seen as a generalization/generification of C's memcpy(). It also maintains the requirement of memcpy(), for the range [first, last) to be disjoint from the range [d_first , d_first + std::distance(first, last)); otherwise we have undefined behavior.

My question: Is there a generic version of std::memmove (i.e. which doesn't make that requirement and would typically be implemented using a temporary buffer)? If not, how come?

Answer 1

C++ doesn't have drop in replacements for memcpy and memmove.

For non-overlapping ranges usually std::copy is used, but std::copy_backwards can also be used.

For overlapping ranges you need to use std::copy or std::copy_backwards depending on the nature of the overlap.

Also for copy-expensive objects std::move (<algorithm>) and std::move_backward respectively can be used instead if the original objects don't need to keep their value.

std::copy

template< class InputIt, class OutputIt >
OutputIt copy( InputIt first, InputIt last, OutputIt d_first );

Copies the elements in the range, defined by [first, last), to another range beginning at d_first.

Copies all elements in the range [first, last) starting from first and proceeding to last - 1. The behavior is undefined if d_first is within the range [first, last). In this case, std::copy_backward may be used instead.

Notes
When copying overlapping ranges, std::copy is appropriate when copying to the left (beginning of the destination range is outside the source range) while std::copy_backward is appropriate when copying to the right (end of the destination range is outside the source range).

std::copy_backward

template< class BidirIt1, class BidirIt2 >
BidirIt2 copy_backward( BidirIt1 first, BidirIt1 last, BidirIt2 d_last );

Copies the elements from the range, defined by [first, last), to another range ending at d_last. The elements are copied in reverse order (the last element is copied first), but their relative order is preserved.

The behavior is undefined if d_last is within (first, last]. std::copy must be used instead of std::copy_backward in that case.

Notes
When copying overlapping ranges, std::copy is appropriate when copying to the left (beginning of the destination range is outside the source range) while std::copy_backward is appropriate when copying to the right (end of the destination range is outside the source range).