Return type forwarding in generic code
For non-generic code, like the initial example you gave, you can manually select to get a reference as a return type:
auto const& Example(int const& i)
{
return i;
}
but in generic code you want to be able to perfectly forward a return type without knowing whether you are dealing with a reference or a value. decltype(auto)
gives you that ability:
template<class Fun, class... Args>
decltype(auto) Example(Fun fun, Args&&... args)
{
return fun(std::forward<Args>(args)...);
}
Delaying return type deduction in recursive templates
In this Q&A a few days ago, an infinite recursion during template instantiation was encountered when the return type of the template was specified as decltype(iter(Int<i-1>{}))
instead of decltype(auto)
.
template<int i>
struct Int {};
constexpr auto iter(Int<0>) -> Int<0>;
template<int i>
constexpr auto iter(Int<i>) -> decltype(auto)
{ return iter(Int<i-1>{}); }
int main() { decltype(iter(Int<10>{})) a; }
decltype(auto)
is used here to delay the return type deduction after the dust of template instantiation has settled.
Other uses
You can also use decltype(auto)
in other contexts, e.g. the draft Standard N3936 also states
7.1.6.4 auto speci?er [dcl.spec.auto]
1 The auto
and decltype(auto)
type-speci?ers designate a placeholder
type that will be replaced later, either by deduction from an
initializer or by explicit speci?cation with a trailing-return-type.
The auto
type-speci?er is also used to signify that a lambda is a
generic lambda.
2 The placeholder type can appear with a function declarator in the decl-specifier-seq, type-specifier-seq,
conversion-function-id, or trailing-return-type, in any context where such a declarator is valid. If the function
declarator includes a trailing-return-type (8.3.5), that specifies the declared return type of the function.
If the declared return type of the function contains a placeholder type, the return type of the function is
deduced from return statements in the body of the function, if any.
The draft also contains this example of variable initialization:
int i;
int&& f();
auto x3a = i; // decltype(x3a) is int
decltype(auto) x3d = i; // decltype(x3d) is int
auto x4a = (i); // decltype(x4a) is int
decltype(auto) x4d = (i); // decltype(x4d) is int&
auto x5a = f(); // decltype(x5a) is int
decltype(auto) x5d = f(); // decltype(x5d) is int&&
auto x6a = { 1, 2 }; // decltype(x6a) is std::initializer_list<int>
decltype(auto) x6d = { 1, 2 }; // error, { 1, 2 } is not an expression
auto *x7a = &i; // decltype(x7a) is int*
decltype(auto)*x7d = &i; // error, declared type is not plain decltype(auto)