How to correctly use std::reference_wrappers
I am trying to understand std::reference_wrapper
.
The following code shows that the reference wrapper does not behave exactly like a reference.
#include <iostream>
#include <vector>
#include <functional>
int main()
{
std::vector<int> numbers = {1, 3, 0, -8, 5, 3, 1};
auto referenceWrapper = std::ref(numbers);
std::vector<int>& reference = numbers;
std::cout << reference[3] << std::endl;
std::cout << referenceWrapper.get()[3] << std::endl;
// I need to use get ^
// otherwise does not compile.
return 0;
}
If I understand it correctly, the implicit conversion does not apply to calling member functions. Is this an inherent limitation? Do I need to use the std::reference_wrapper::get
so often?
Another case is this:
#include <iostream>
#include <functional>
int main()
{
int a = 3;
int b = 4;
auto refa = std::ref(a);
auto refb = std::ref(b);
if (refa < refb)
std::cout << "success" << std::endl;
return 0;
}
This works fine, but when I add this above the main
definition:
template <typename T>
bool operator < (T left, T right)
{
return left.someMember();
}
The compiler tries to instantiate the template and forgets about implicit conversion and the built in operator.
Is this behavior inherent or am I misunderstanding something crucial about the std::reference_wrapper
?
Solution 1:
Class std::reference_wrapper<T>
implements an implicit converting operator to T&
:
operator T& () const noexcept;
and a more explicit getter:
T& get() const noexcept;
The implicit operator is called when a T
(or T&
) is required. For instance
void f(some_type x);
// ...
std::reference_wrapper<some_type> x;
some_type y = x; // the implicit operator is called
f(x); // the implicit operator is called and the result goes to f.
However, sometimes a T
is not necessarily expected and, in this case, you must use get
. This happens, mostly, in automatic type deduction contexts. For instance,
template <typename U>
g(U x);
// ...
std::reference_wrapper<some_type> x;
auto y = x; // the type of y is std::reference_wrapper<some_type>
g(x); // U = std::reference_wrapper<some_type>
To get some_type
instead of std::reference_wrapper<some_type>
above you should do
auto y = x.get(); // the type of y is some_type
g(x.get()); // U = some_type
Alternativelly the last line above could be replaced by g<some_type>(x);
.
However, for templatized operators (e.g. ostream::operator <<()
) I believe you can't explicit the type.