Template argument deduction for references as arguments

问题

I am trying to profoundly understand Template Argument Deduction. One point I am not understanding is, how I should apply the rules in the standard here for the types A and P for the following case (there is sadly no example on cppreference.com, see below the relevant section)

template<typename T>
void foo(T t);

void call_with_reference(int& r) {
    foo(r)
}

• P is no reference typ:
  which gives P := T
• A := int&

-> Match P and A which gives: T is deduced to int&

which is cleary wrong. Where is the rule in the standard that says references from A are removed? A non-confusing, unambiguous clear answer would be very much appreciated.

Relevant Section:

回答1:

A is the type of an expression. Expression type is described by [expr.type]/1:

If an expression initially has the type “reference to T” ([dcl.ref], [dcl.init.ref]), the type is adjusted to T.

So here A is int.

This expression is an lvalue but that will not play any role since P is not a reference.

来源：https://stackoverflow.com/questions/56548276/template-argument-deduction-for-references-as-arguments