why iterating over unordered_map with unique_ptr forces a pair with a key that is const? - for-loop

Question

I found a compiler error that I can not understand why it happens only for std::map with std::unique_ptr.
Assume we have the follow unoredered map object and an iteration code for it:
std::unordered_map<uint32_t, std::shared_ptr<char>> map;
for (const std::pair<uint32_t, std::shared_ptr<char>>& item : map)
{
// do something
}
That's compiled well, but if we use unique pointer instead the shared pointer as follows, then we get a compiler error about the iteration pair type:
std::unordered_map<uint32_t, std::unique_ptr<char>> map;
for (const std::pair<uint32_t, std::unique_ptr<char>>& item : map)
{
// do something
}
error C2440: 'initializing': cannot convert from 'std::pair<const _Kty,_Ty>' to
'const std::pair<uint32_t,std::unique_ptr<char,std::default_delete<_Ty>>>
Following this error, we can just add 'const' to the key type and it is compiled.
for (const std::pair<const uint32_t, std::unique_ptr<char>>& item : map)
^^^
|||
Why is this compiling error happens just for unique pointer?

Answer 1

A friend gave me the answer and I'm sharing.
The answer lies in the concept of implicit conversion(copy) a compiler is allowed to do.
Lets see the follow simple example:
const char x = 4;
const char& r = x; // r is reference for x. Checking their addresses yields the same
// address.
const int& ir = c; // implicit creation of object + conversion(copy).
// ir is different type of x, therefor compiler does implicit
// conversion(copy): it creates behind the scene an object of int,
// convert x into this temporary object. The temporary int object is
// then bound to the reference ir. Checking addresses of ir and x
// yields different addresses because ir is reference of the temporary
// object, not to x
So even we use reference - planning for just pointing on an existing object,
we may actually have an object construction + copy (if the types are different and there is conversion between them).
The same happens in the loop I gave in the question:
std::unordered_map<uint32_t, std::shared_ptr<char>> map;
for (const std::pair<uint32_t, std::shared_ptr<char>>& item : map)
{
// do something
}
while the real object held by the map is of type of
std::pair<const uint32_t, std::shared_ptr<char>>
the loop uses reference of different type:
std::pair<uint32_t, std::shared_ptr<char>>
and therefor, behind the scene, for every iteration, a new temporary object is implicitly constructed and a copy operation for conversion is done.
Not only that it isn't efficient, but it also failed to be compiled when the value is unique_ptr because the conversion does copy and unique_ptr can not be copied.
Thats why using 'auto' may save you such mistakes,
std::unordered_map<uint32_t, std::shared_ptr<char>> map;
for (const auto& item : map)
{
// do something
}
I myself sometimes prefer to play and use the explicit form instead 'auto' in order to face such problems and learn :)