compile error with std::reference_wrapper "has no member named" when used with class

Question

So I've found this fancy stuff std::reference_wrapper in C++11 which I found could be useful in future program development. I played with it for a bit and I thought it could replace good-old reference in some cases for better usage, until I bumped into the following error:

#include <iostream>
#include <functional>

class testbench {
public:
    int ownerid = 5;
    bool camper = false;
};

class testing {
public:
    testing(testbench &x);
    int quack = 3;
    std::reference_wrapper <testbench> camper;
};

testing::testing(testbench &x):camper(x) {
}

int main() {
    testbench tempo;
    testing test_a (tempo);
    std::cout << test_a.camper.ownerid;
}

this would cause an error in gcc (I'm using Code::Blocks as Editor and gcc as compiler)

'class std::reference_wrapper<testbench>' has no member named 'ownerid'

If the line

    std::reference_wrapper <testbench> camper;

is changed to :

    testbench &camper;

Then everything is worked as I expected.

How do I fix this, or is it just my misunderstanding std::reference_wrapper as a general improvement of reference?

Cast them explicitly. "There is no additional level of syntactical indirection. Pointers have to be dereferenced to obtain an lvalue to the object they refer to; reference_wrappers have an implicit conversion operator and can be called like the object they wrap." - source. — user202729
– user202729, Commented Feb 27, 2018 at 3:32

Stephen Newell · Accepted Answer · 2018-02-27 03:37:01Z

7

Your understanding is wrong. std::reference_wrapper can implicitly cast to a T& (in your case, testbench) but when you use the member accessor (test_a.camper.ownerid) it's operating on the camper, not the object it's wrapping.

You can either do a cast or use the get member function. This is untested, but I think it'll work:

test_a.camper.get().ownerid;

answered Feb 27, 2018 at 3:37

Stephen Newell

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