Reference initialization with constructor

Question

const T   a {}; // constant of type T
const T&  b {}; // ???
      T   c {}; // variable of type T
      T&  d {}; // error

What is the difference between a and b?
b is a reference but I don't assign a object to it: in that instruction I initialize it by T constructor.
The address of b is between the addresses of a and c, so it seems the b and a have no differences.

And if I can declare and initialize b why d gives compilation error?

I talked about a generic type T. I tested the code above either for primitive types and classes and the results are the same.

a and b are exactly the same except for the result of decltype(x) — M.M
– M.M, Commented Mar 19, 2020 at 23:30

Vlad from Moscow · Accepted Answer · 2020-03-19 21:18:28Z

2

In these declarations

const T   a {}; // constant of type T
const T&  b {};

there is created constant object a that is default initialized and constant reference b to temporary object that is default initialized.

The compiler issues an error for this declaration of a reference

T&  d {}; // error

because there is declared a non-constant reference to a temporary object.

You could declare an rvalue reference the following way

T&& d {};

Here is a demonstrative program

#include <iostream>

int main() 
{
    const int &ri {};

    std::cout << "ri = " << ri << '\n';

    int && rri {};

    std::cout << "rri = " << rri << '\n';

    return 0;
}

The program output is

ri = 0
rri = 0

edited Mar 19, 2020 at 21:18

answered Mar 19, 2020 at 21:12

Vlad from Moscow

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2 Comments

DPD- Over a year ago

So I guess that in b the lifetime of the temporary object is extended and then to "correct" d I would have to use an rvalue reference: T&& d {}; Right?

Vlad from Moscow Over a year ago

@DPD- The lifetime of the both temporary objects is as long as the lifetime of references thet refer these temporary objects.

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