c++ - `auto` specifier type deduction for references

Question

Let's consider the following code snippet

void Test()
  {
  int x = 0;

  int& rx = x;
  int* px = &x;

  auto apx = px;    // deduced type is int*
  auto arx = rx;    // deduced type is int
  }

One could draw an analogy from pointer types expecting that the deduced type of arx is int&, but it is int in fact.

What is the rule in Standard which governs that? What is the reason behind it? Sometimes I get caught by it in a case like this:

const BigClass& GetBigClass();
...
auto ref_bigclass = GetBigClass();   // unexpected copy is performed

Answer 1

Use auto&:

auto& ref_bigclass = GetBigClass();

References are supposed to be transparent: any operation on them happens on the object they refer to, there is no way to 'get' reference itself.

UPD: This is covered in 7.1.6.4/6:

Once the type of a declarator-id has been determined according to 8.3, the type of the declared variable using the declarator-id is determined from the type of its initializer using the rules for template argument deduction.

And template argument deduction is defined in 14.8.2.1/3:

If template parameter type P is a reference type, the type referred to by P is used for type deduction.

P.S. Note that this is different for decltype: decltype(rx) will yield int& type (7.1.6.2/4).