Generally, is dereference pointer expression results a reference type?

Question

Deferencing pointer leads to using the value of the object indirectly. But I've never really understood what does the "using" means. I started to think the question until my compiler yield an error for the following code

int i = 0, *pi = &i;
decltype(*pi) c; // error: 'c' declared as reference but not initialized.

I looked at the error for a very long time and searched some questions I can only give out the following arguments. I don't know if they are correct or not.

Arguments 1:

1) *p is an expression that is not a variable (or non-variable expression)

2) dereferencing pointer expression yields a reference, we are in fact using a reference to access the value of the object

Arguments 2:

the dereferencing expression only for which decltype returns a reference, it is not a general case

Please points out any incorrectness or inaccurate descriptions of the above arguments.

Answer 1

Dereferencing a pointer yields an lvalue expression of the pointed-to type designating the object or function pointed to. It does not yield a reference.^**pi is an lvalue of type int.

decltype (with an exception not relevant here) reports both an expression's type and its value category, the latter being encoded with reference types. Since *pi is an lvalue, it's encoded as an lvalue reference type, so decltype(*pi) is int &: int for the type, & for the value category.

Expressions never have reference type because any referenceness is adjusted away "prior to any further analysis".

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_{^* This isn't a merely technical distinction: per the direction of core issue 232 and core issue 453, there are valid dereference expressions you can write where binding its result to a reference would cause undefined behavior.}