Quick A: To enabling moving out of *iterator.

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What is move_iterator for?

If I understand it correct, a=std::move(b) binds reference a to the address of b. And after this operation the content that b points to is not guaranteed.

The implementation of move_iterator here has this line

auto operator[](difference_type n) const -> decltype(std::move(current[n]))
  { return std::move(current[n]); }

However, I don't think it makes sense to std::move an element in an array. What happens if a=std::move(b[n])?

The following example confuses me also:

std::string concat = std::accumulate(
                             std::move_iterator<iter_t>(source.begin()),
                             std::move_iterator<iter_t>(source.end()),
                             std::string("1234"));

Since the concat will itself allocate a continuous chunk of memory to store the result, which will not have any overlap with source. The data in source will be copied to concat but not moved.