Const in auto type deduction

Question

I am reading Effective modern C++ from Scott Meyers. Item 1 contains the following example:

template<typename T>
void f(T& param);       // param is a reference
int x = 27;             // x is an int
const int cx = x;       // cx is a const int
f(cx);                  // T is const int,
                        // param's type is const int&

In Item 3 appears the following example:

Widget w;
const Widget& cw = w;
auto myWidget1 = cw;             // auto type deduction:
                                 // myWidget1's type is Widget

Based on Item 1 I expected myWidget1's type to be const Widget. Am I missing something?

Answer 1

In most cases auto follows the rules of template argument deduction:

§ 7.1.6.4 [dcl.spec.auto]/p6:

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. Let T be the type that has been determined for a variable identifier d. Obtain P from T by replacing the occurrences of auto with either a new invented type template parameter U or, if the initializer is a braced-init-list (8.5.4), with std::initializer_list<U>. The type deduced for the variable d is then the deduced A determined using the rules of template argument deduction from a function call (14.8.2.1).

§ 14.8.2.1 [temp.deduct.call]/p2:

If P is not a reference type:

• [...]

• If A is a cv-qualified type, the top level cv-qualifiers of A's type are ignored for type deduction.

If you want myWidget1 to be of type const Widget&, it should be declared as a reference type, e.g.:

auto& myWidget1 = cw;
//  ^

DEMO