Why I can declare a const reference using type-alias?

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Category:Languages

I have a simple question: As I know I can declare a const pointer to some datatype or a pointer to constant datatype but I can only declare a reference to a constant datatype only and no constant reference to a datatype; the fact that a reference is already constant because it cannot be rebound to another object.

So when I try to create a const ref to someDataType I get compile-time error. But the thing that matters me is when used with type alias using typedef or using. e.g:

#include <iostream>  int main() {      int i{ 10 };     //  int& const r1{ i }; // error: ‘const’ qualifiers cannot be applied to ‘int&’. Ok here.     using rInt = int&; // or typedef int& rInt;     const rInt r2{ i }; // why const is allowed here?     ++r2; // this proves that the const is applied to the reference not to the object referred to.      std::cout << r2 << std::endl; // 11  } 

As you can see above I can add const to the reference which I think is Redundant in that context. But why C++ allows this with type-aliases but not directly?

 


Because the standard say so:

[dcl.ref] ... Cv-qualified references are ill-formed except when the cv-qualifiers are introduced through the use of a typedef-name ([dcl.typedef], [temp.param]) or decltype-specifier ([dcl.type.simple]), in which case the cv-qualifiers are ignored

This is similar to how you cannot declare a reference reference, while it is possible through a typedef (where the references collapse into one):

int i; int& iref = i; //int& & irefref = iref; // not OK using Iref = int&; Iref& iretypedef = iref; // OK; collapses into int& 

The CV-collapsing rules, just like reference collapsing rules are essential to make templates and type deductions usable.

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