In this example, they mean the exact same thing.
However, there are a few advantages to using the trailing return type form consistently (Phil Nash calls these "East End Functions", since the return type is on the east end).
Using parameters. Obviously when using parameters to determine the return type, you must use a trailing return type.
template <typename T>
auto print(T const& t) -> decltype(std::cout << t) { return std::cout << t; }
Name lookup. In a trailing return type, name lookup includes the class scope for member function definitions. This means you don't have to retype the class if you want to return a nested class:
Type C::foo() { ... } // error: don't know what Type is
C::Type C::foo() { ... } // ok
auto C::foo() -> Type { ... } // ok
Likewise, for defining member functions where the class name for some reason must be disambiguated to be in the global namespace and the return type is a class:
D ::C::foo() { ... } // error, parsed as D::C::foo() { ... }
auto ::C::foo() -> D { ... } // ok
There are cases where trailing-return-type is mandatory, there are cases where it is helpful, and there are cases where it does the same thing. There are not cases where it is worse for reasons other than simply character count.
Plus, mathemtically we're used to thinking of functions as A -> B
and not so much B(A)
, and so auto(*)(A) -> B
as a function pointer taking an A
and returning a B
is a little closer to that view than B(*)(A)
.
On the other hand, writing auto main() -> int
looks ridiculous.
But honestly, that's mostly because of unfamiliarity. There's nothing inherently ridiculous about it. If anything, it's a bit unfortunate that the language uses auto
here as a way to declare a function - not because it's too long (i.e. some other languages use fun
or even fn
) - but because it's not really distinct from other uses of auto
. If it were func
instead, I think it would have been better (although it would not make any sense to change now).
Ultimately, this is purely opinion based. Just write code that works.
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