Is converting decltype's expression from constant l-value into an r-value discards `const` too?

Question

Here I want to know how type specifier decltype works:

const int& rci = 5;// const ref bound to a temporary
decltype (rci) x = 2;
decltype (rci + 0) y = 10;
// ++x; // error: increment of read-only reference ‘x’|
++y; // why does this work

std::cout << typeid(x).name() << std::endl; // i
std::cout << typeid(y).name() << std::endl; // i

• Why y has only type int rather than const int?

• Does converting a constant lvalue into an rvalue discards in addition to reference operator, const qualifier too?

• Why typeid shows that the type of x is just i but not const int&?

Answer 1

> Why y has only type int rather than const int?

Because (rci + 0) is a prvalue, and prvalues of primitive types have const stripped.

> Does converting a constant lvalue into an rvalue discards in addition to reference operator, const qualifier too?

For non-class types, const qualifier is discarded. From the standard:

7.3.1.1
[...] If T is a non-class type, the type of the prvalue is the cv-unqualified version of T.

> Why typeid shows that the type of x is just i but not const int&?

From cppreference:

In all cases, cv-qualifiers are ignored by typeid (that is, typeid(const T) == typeid(T)).