C++ How does an std::reference_wrapper implicitly convert to reference?

Question

I've recently begun using the std::reference_wrapper class. When replacing certain uses of primitive references, I noticed that I did not have to use the get() function to pass the reference_wrappers as parameters to functions that take a normal reference.

void foo(const T& a);
//...
T obj;
std::reference_wrapper<const T> ref(obj);
foo(ref);  //works!
//foo(ref.get()); also works, but I expected that it would be required

How does std::reference_wrapper implicitly convert to a primitive reference when it is passed into a function?

Jerry Coffin · Accepted Answer · 2012-11-01 04:52:11Z

8

reference_wrapper includes (§20.8.3):

// access
operator T& () const noexcept;

This is a conversion operator, and since it isn't specified as explicit, it allows implicit conversion from reference_wrapper<T> to T&.

answered Nov 1, 2012 at 4:52

Jerry Coffin

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Pubby · Accepted Answer · 2012-11-01 23:52:24Z

5

It overloads this conversion operator:

operator T& () const;

http://en.cppreference.com/w/cpp/utility/functional/reference_wrapper/get

edited Nov 1, 2012 at 23:52

answered Nov 1, 2012 at 4:51

Pubby

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