l-value specifies const object when using std::make_pair - c++

Question

struct MapInserter
{
private:
int count;
public:
explicit MapInserter()
: count(0)
{
}
std::pair<int, std::string> operator()(std::string& value)
{
return std::make_pair(count++, value);
}
};
vector<std::string> words = { "one", "two", "three","four","five" };
std::map<int, std::string> map;
MapInserter inserter;
transform(words.begin(), words.end(), map.begin(), inserter);
for (auto it = map.begin(), end = map.end(); it != end; ++it)
cout << it->first << " : " << it->second << endl;
return 0;
that's the code. VS returns a compile error regarding l-value specifies const object.
Clicking the error moves you to the following code in a file named utility
template<class _Other1,
class _Other2>
_Myt& operator=(pair<_Other1, _Other2>&& _Right)
{ // assign from moved compatible pair
first = _STD forward<_Other1>(_Right.first);
second = _STD forward<_Other2>(_Right.second);
return (*this);
}
at first, i've had the operator() take const std::string& so I removed the const, because it's obviously talking about the make_pair function. But it still hasn't gone away. Can anyone point me to what this error is about?

Answer 1

The problem is that std::transform() will try to assign to existing elements of the target container. Keys of a map are constant and cannot be assigned to, which is why you're getting a compiler error. But even if they were, you'd get undefined behavior at run-time here, because the target container is empty, and std::transform() would expect it to contain as many elements as the input range.
You should use std::inserter() to create an inserter iterator, like so:
vector<std::string> words = { "one", "two", "three","four","five" };
std::map<int, std::string> map;
MapInserter inserter;
transform(words.begin(), words.end(), std::inserter(map, map.begin()), inserter);
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Here is a live example.
Moreover, taking the value string by mutable lvalue reference in the call operator of your MapInserter is not a good idea: you don't want the argument to be modified, so you should either take it by const& or - my advice - take it by value and then move it into the returned pair, like so:
std::pair<int, std::string> operator()(std::string value)
{
return {count++, std::move(value)};
}
Since std::pair's constructor is not explicit, you do not even need the call to std::make_pair() in this case.