What is move_iterator for

Question

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.

Answer 1

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.

Ah, no: a is not necessarily a reference. The above use of std::move also grants the compiler permission to call decltype(a)::operator=(decltype(b)&&) if it exists: such assignment operators are used when during the assignment to a the value of b need not be preserved, but b must still be left in some sane state for destruction.

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

It can make sense... it just means that each array elements may be efficiently assigned/moved to another variable, but only once per element. After they've been moved-from, a properly-written move constructor or assignment operator should leave objects in a valid but unspecified state, which means you'd usually want to set them again before reading from them.

My answer here shows how someone could append/move elements from a list to a vector. With current C++ Standards, you can create move_iterators directly like that.

The code below shows how - even with older compilers / C++ Standards - make_move_iterator can be used with std::copy if you want to move from the elements in the source iterator range.

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>

struct X
{
    X(int n) : n_(n) { }
    X(const X& rhs) : n_(rhs.n_) { }
    X(X&& rhs) : n_{ rhs.n_ } { rhs.n_ *= -1; std::cout << "=(X&&) "; }
    X& operator=(X&& rhs) { n_ = rhs.n_; rhs.n_ *= -1; std::cout << "=(X&&) "; return *this; }
    int n_;
};

int main()
{
    std::vector<X> v{2, 1, 8, 3, 4, 5, 6};
    std::vector<X> v2{};

    std::copy(v.begin() + 2, v.end(), std::insert_iterator(v2, v2.end()));
    for (auto& x : v)
        std::cout << x.n_ << ' ';
    std::cout << '\n';

    std::copy(std::make_move_iterator(v.begin() + 2), std::make_move_iterator(v.end()), std::insert_iterator(v2, v2.end()));
    for (auto& x : v)
        std::cout << x.n_ << ' ';
    std::cout << '\n';
}

Output:

2 1 8 3 4 5 6 
=(X&&) =(X&&) =(X&&) =(X&&) =(X&&) 2 1 -8 -3 -4 -5 -6 

Code can be run / edited on coliru.