How do reference_wrapper and std::ref work?

Question

I am trying to understand how std::ref works.

#include <functional>
#include <iostream>

template <class C>
void func(C c){
    c += 1;
}

int main(){
    int x{3};
    std::cout << x << std::endl;
    func(x);
    std::cout << x << std::endl;
    func(std::ref(x));
    std::cout << x << std::endl;
}

Output : 3 3 4

In the code above, I think the template parameter C for the third function call is instantiated as std::reference_wrapper<int>. While reading the reference, I noticed there is no += operator in std::reference_wrapper<int>. Then, how is c += 1; valid?

Mike Seymour · Accepted Answer · 2013-05-14 11:51:53Z

19

how is c += 1; valid?

Because reference_wrapper<int> is implicitly convertible to int& via its conversion operator; and implicit conversions are considered for operands if there is no suitable overload for the operand type itself.

answered May 14, 2013 at 11:51

Mike Seymour

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2 Comments

Mike Seymour Over a year ago

@Sungmin: Indeed, c itself can't change type; but += is applied to the result of c.operator int&() which will be a reference to target of the reference_wrapper.

Louis Semprini Over a year ago

Note this magic only works for operators. For example, if x were of class type and had a method x.foo(), and you tried to wrap a reference to x in a reference_wrapper rx, you could not just say rx.foo(). You would have to say rx.get().foo(), which is about as ugly as just using a pointer px->foo. So unfortunately std::reference_wrapper is not a magical do-all wrapper for references (which is what a lot of people hope to use it for, e.g. for copying objects with reference members).

juanchopanza · Accepted Answer · 2013-05-14 11:52:12Z

12

std::reference_wrapper<T> has a conversion operator to T&. This means that std::reference_wrapper can be implicitly converted to int& in your example.

answered May 14, 2013 at 11:52

juanchopanza

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2 Comments

Sungmin Over a year ago

The two answers were uploaded almost the same time. But Mike Seymour's answer is a little bit faster. Apart from that, I cannot determine which one to pick. Sorry for that. Thanks anyway.

juanchopanza Over a year ago

@Sungmin You are welcome. I think Seymour's answer is better anyway: he explains about the operators overloads and operands, which I failed to do.

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