What is the result of decltype(“Hello”)?

[Note: Originally, this was not meant to be a self-answered question; I just happened to find the answer myself while I was describing my attempts to investigate, and I thought it would have been nice to share it.]

According to Annex C (2.14.5) of the C++11 Standard:

The type of a string literal is changed from “array of char” to “array of const char.” [….]

Moreover, Paragraph 7.1.6.2/4 specifies (about the result of decltype):

The type denoted by decltype(e) is defined as follows:

— if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;

— otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;

— otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

— otherwise, decltype(e) is the type of e.

Since string literals are lvalues, according to the above Paragraph and the Paragraph from Annex C, the result of decltype("Hello") is an lvalue reference to an array of size 6 of constant narrow characters:

#include <type_traits>

int main()
{
    // This will NOT fire
    static_assert(
        std::is_same<decltype("Hello"), char const (&)[6]>::value, 
        "Error!"
        );
}

Finally, even though the hello variable is also an lvalue, the second compile-time assertion from the question’s text does not fire, because hello is an unparenthesized id-expression, which makes it fall into the first item of the above list from Paragraph 7.1.6.2/4. Therefore, the result of decltype(hello) is the type of the entity named by hello, which is char const[6].