When does `auto` in C 11 deduce a value or a reference?

The Subtleties of C 11 "auto" Semantics

In C 11, the auto keyword offers a convenient way to automatically deduce the type of a variable based on its initializer. However, there are certain нюансы to consider regarding whether auto will resolve to a value or a reference.

Understanding the Rule

The rule is straightforward: it depends on how the variable is declared.

<code class="cpp">int i = 5;
auto a1 = i;    // value
auto &a2 = i;  // reference</code>

In the first case, a1 is a value because it was not declared as a reference. In the second case, a2 is a reference because it was explicitly declared as one.

Clarifying the Uncertainties

Let's apply this rule to the examples you provided:

• auto p = get_foo();: get_foo() returns a reference to a smart pointer. Since p is declared without a reference, it will be a copy, resulting in a reference type object.
• auto sp = s_foo;: s_foo is a static reference to a smart pointer. When assigning it to sp, the reference will be copied, resulting in a reference type object.
• for (auto foo: c): foo will be a copy of each iterator in the vector because it is not declared as a reference.

Proof with Template Metaprogramming

The following code demonstrates this behavior using template metaprogramming:

<code class="cpp">#include <typeinfo>
#include <iostream>

template
struct A
{
    static void foo(){ std::cout 
struct A
{
    static void foo(){ std::cout ::foo();       // value
    A<decltype>::foo();       // reference
    A<decltype>::foo();      // value
    A<decltype>::foo();      // value
    A<decltype>::foo();   // reference
    A<decltype>::foo();      // value
}</decltype></decltype></decltype></decltype></decltype></iostream></typeinfo></code>

Output:

value
reference
value
value
reference
value

This confirms that the type of auto is determined by its declaration, not by the type of its initializer.

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