Understanding C 11 "auto" Semantics
In C 11, the "auto" keyword allows for automatic type deduction. However, determining whether it resolves to a value or a reference can sometimes be ambiguous.
Type Deduction Rules:
The rule for type deduction in "auto" is straightforward: it is based on how the object is declared.
For instance:
<code class="cpp">int i = 5;</code>
If we declare:
<code class="cpp">auto a1 = i; // value auto &a2 = i; // reference</code>
a1 will be deducted as a value type (int), while a2 will be deducted as a reference type (int&).
Ambiguous Cases:
Consider the following examples:
<code class="cpp">const std::shared_ptr<foo>& get_foo(); auto p = get_foo(); // Copy or reference?</foo></code>
get_foo returns a reference, but p is declared as an "auto." In this case, auto will deduce the type of p to be a reference to Foo.
<code class="cpp">static std::shared_ptr<foo> s_foo; auto sp = s_foo; // Copy or reference?</foo></code>
s_foo is a static instance of a shared pointer to Foo. sp is declared using "auto," so it will be deducted as a shared pointer to Foo.
<code class="cpp">std::vector<:shared_ptr>> c; for (auto foo: c) { // Copy for every loop iteration?</:shared_ptr></code>
Here, "auto" will iterate through the vector, deducing the type of foo to be a shared pointer to Foo. This means that a new temporary shared pointer is created for each iteration.
Demonstration:
The following code demonstrates these concepts:
<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
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