Are std::move and std::copy identical?

问题

I tried to do something like:

std::copy(std::make_move_iterator(s1.begin()), std::make_move_iterator(s1.end()), 
          std::make_move_iterator(s2.begin()));

And got this error:

error: using xvalue (rvalue reference) as lvalue
        *__result = std::move(*__first);

Which seemed confusing to me. The same thing happens if you use std::move. It appears that GCC internally uses a function called std::__copy_move_a which moves rather than copies. Does it matter whether you use std::copy or std::move?

---

#include <string>
#include <iostream>
#include <algorithm>
#include <iterator>
#include <cstring>

struct Test
{
    typedef std::string::value_type value_type;
    std::string data;

    Test()
    {
    }

    Test(const char* data)
        : data(data)
    {
    }

    ~Test()
    {
    }

    Test(const Test& other)
        : data(other.data)
    {
        std::cout << "Copy constructor.\n";
    }

    Test& operator=(const Test& other)
    {
        data = other.data;
        std::cout << "Copy assignment operator.\n";
        return *this;
    }

    Test(Test&& other)
        : data(std::move(other.data))
    {
        std::cout << "Move constructor.\n";
    }

    decltype(data.begin()) begin()
    {
        return data.begin();
    }

    decltype(data.end()) end()
    {
        return data.end();
    }

    void push_back( std::string::value_type ch )
    {
        data.push_back(ch);
    }
};

int main()
{
    Test s1("test");
    Test s2("four");
    std::copy(std::make_move_iterator(s1.begin()), std::make_move_iterator(s1.end()), 
              std::make_move_iterator(s2.begin()));
    std::cout << s2.data;
}

回答1:

std::move(a, b, c); is semantically identical to

std::copy(std::make_move_iterator(a),
          std::make_move_iterator(b),
          c);

Your efforts to use them both failed because the third argument - the output iterator - should not be a move iterator. You are storing into the third iterator, not moving from it. Both

std::copy(std::make_move_iterator(s1.begin()),
          std::make_move_iterator(s1.end()),
          s2.begin());

and

std::move(s1.begin(), s1.end(), s2.begin());

should do what you want.

回答2:

std::move moves the elements if possible, and copies otherwise. std::copy will always copy.

libstdc++'s copy_move_a also takes a template parameter _IsMove. That, and the iterator types, it delegates to a __copy_move class template that is partially specialized for different iterator categories, etc. but most importantly: Whether to move or not. One of the specializations is

#if __cplusplus >= 201103L
  template<typename _Category>
    struct __copy_move<true, false, _Category>
    // first specialized template argument is whether to move
    {
      template<typename _II, typename _OI>
        static _OI
        __copy_m(_II __first, _II __last, _OI __result)
        {
      for (; __first != __last; ++__result, ++__first)
        *__result = std::move(*__first); // That may be your line
      return __result;
    }
    };
#endif

---

Your code fails to compile for a completely different reason: The second range is given through move_iterators. If you dereference them, they return an rvalue reference to the object - and you can't assign something to an xvalue of scalar type.

int i;
std::move(i) = 7; // "expression not assignable" -- basically what your code does

The std::move is implicitly included in *__result and is of the same value category, that is, an xvalue.

For your example,

std::copy(std::make_move_iterator(s1.begin()), std::make_move_iterator(s1.end()),
          s2.begin());

should work fine.

来源：https://stackoverflow.com/questions/26432990/are-stdmove-and-stdcopy-identical