Is it possible to return an object of type T by reference from a lambda without using trailing return type syntax?

Question

Given the following code snippet:

struct T {};   std::function<T&(T&)> f = [](T& obj) -> T& { return obj; }; 

I was wondering if it is possible to infer the correct lambda return type (i.e. T&) without using trailing return type syntax.

Obviously, if I remove -> T& then a compile-time error will occur in that the deduced type would be T.

Answer 1

In C++14, you can use [](T& obj) -> decltype(auto) { return obj; }. In that case, the return type of f is deduced from the declared type of obj (i. e. T& in this case).

Answer 2

No, but in C++14 you could use auto& as the trailing-return-type. If it's typing you're worried about, and compiler upgrades don't worry you at all, then this mostly solves your problem.