In a lambda, how reference is being captured by value

Question

If a variable of reference type is captured in a lambda by value, is the referenced object copied or is it captured by reference?

Small sample with question:

#include <iostream>  struct Test {   int a; };  void testFunc(const Test &test) {   auto a = [=] {     // is 'test' passed to closure object as a copy     // or as a reference?     return test.a;   } ();   std::cout << a; }  int main() {   Test test{1};   testFunc(test); } 

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?