When Does \'auto\' Deduce a Value vs. a Reference in C 11?

C 11 "auto" Semantics

In C 11, the "auto" keyword allows for the compiler to automatically deduce the type of a variable. Understanding the rules for this deduction is crucial when determining whether "auto" will result in a value or a reference type.

Type Deduction for Value and Reference

The primary principle governing type deduction is that the type of "auto" is determined by the way it is declared. For instance:

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

In the above example, "a1" is deduced as an integer value, while "a2" is deduced as an integer reference.

Examples

1. Returning a Reference:

<code class="cpp">const std::shared_ptr<Foo>& get_foo();
auto p = get_foo();</code>

In this case, "auto" deduces a reference type because "get_foo()" returns a const reference.

2. Static Pointer:

<code class="cpp">static std::shared_ptr<Foo> s_foo;
auto sp = s_foo;</code>

Here, "auto" deduces a value type because "s_foo" is a static pointer, which is not a reference.

3. Looping Over a Container of Pointers:

<code class="cpp">std::vector<std::shared_ptr<Foo>> c;
for (auto foo : c) {</code>

In this loop, "auto" ensures that "foo" is deduced as a shared pointer to Foo. Each iteration creates a value copy of the pointer.

Conclusion

To summarize, the type deduction rules for "auto" in C 11 are straightforward. If the initialization expression is a value, "auto" deduces a value type. If the initialization expression is a reference, "auto" deduces a reference type. Understanding these rules is essential for effective use of "auto" and avoiding unexpected behavior.

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