User-defined conversion operator doesn't work on references - c++

Question

I have a simple primitive type wrapper:
template <typename T>
class Scalar {
public:
explicit Scalar(T value) : value{value} {}
Scalar(Scalar&& other) = default;
Scalar& operator=(Scalar&& other) = default;
Scalar(const Scalar& other) = default;
Scalar& operator=(const Scalar& other) = default;
template <typename U>
explicit operator Scalar<U>() {
return Scalar<U>{static_cast<U>(this->value)};
}
inline T getValue() const noexcept { return this->value; }
private:
T value;
};
Casting Scalar values works well, but somehow it fails for references, e.g.
auto a = Scalar<double>{2.54};
Scalar<int> b = static_cast<Scalar<int>>(a); // works
const auto& c = a;
Scalar<int> d = static_cast<Scalar<int>>(c); // fails
Here's the compilation error (can be checked here http://rextester.com/GOPYU13091), any ideas what might be the issue here?
source_file.cpp: In function ‘int main()’:
source_file.cpp:31:47: error: no matching function for call to ‘Scalar(const Scalar<double>&)’
Scalar<int> d = static_cast<Scalar<int>>(c);
^
source_file.cpp:12:3: note: candidate: Scalar<T>::Scalar(const Scalar<T>&) [with T = int] <near match>
Scalar(const Scalar& other) = default;
^
source_file.cpp:12:3: note: conversion of argument 1 would be ill-formed:
source_file.cpp:31:47: error: could not convert ‘c’ from ‘const Scalar<double>’ to ‘const Scalar<int>&’
Scalar<int> d = static_cast<Scalar<int>>(c);
^
source_file.cpp:10:3: note: candidate: Scalar<T>::Scalar(Scalar<T>&&) [with T = int] <near match>
Scalar(Scalar&& other) = default;
^
source_file.cpp:10:3: note: conversion of argument 1 would be ill-formed:
source_file.cpp:31:47: error: could not convert ‘c’ from ‘const Scalar<double>’ to ‘Scalar<int>&&’
Scalar<int> d = static_cast<Scalar<int>>(c);
^

Answer 1

This is a const vs non-const issue, not a reference vs object issue.
Using
auto& c = a;
Scalar<int> d = static_cast<Scalar<int>>(c);
works for me.
By changing the user defined conversion operator to const member function
template <typename U>
explicit operator Scalar<U>() const {
return Scalar<U>{static_cast<U>(this->value)};
}
also makes sure that the following works.
const auto& c = a;
Scalar<int> d = static_cast<Scalar<int>>(c);