In a lambda, how reference is being captured by value

Question

If the variable of reference-type is being captured in a lambda by value, is it being captured by reference or by value?

Small sample with question:

#inclu         


        

Answer 1

By value. Compilable example:

class C
{
public:
    C()
    {
        i = 0;
    }

    C(const C & source)
    {
        std::cout << "Copy ctor called\n";
        i = source.i;
    }

    int i;
};

void test(C & c)
{
    c.i = 20;

    auto lambda = [=]() mutable {

        c.i = 55;
    };
    lambda();

    std::cout << c.i << "\n";
}

int main(int argc, char * argv[])
{
    C c;
    test(c);

    getchar();
}

Result:

Copy ctor called
20

I guess, that this paragraph of the C++ standard applies:

5.1.2 Lambda expressions

(...) 14. An entity is captured by copy if it is implicitly captured and the capture-default is = or if it is explicitly captured with a capture that does not include an &. For each entity captured by copy, an unnamed nonstatic data member is declared in the closure type. The declaration order of these members is unspecified. The type of such a data member is the type of the corresponding captured entity if the entity is not a reference to an object, or the referenced type otherwise. [ Note: If the captured entity is a reference to a function, the corresponding data member is also a reference to a function. —end note]

That actually makes sense - if local variables are passed by value and parameter passed by reference "acts" as a local variable in function, why would it be passed by reference instead of value?