What does the -> notation after the function parenthesis signify?

Question

For example, in:

  virtual auto create_obj() -> std::unique_ptr<Base>
  {
    return std::unique_ptr<Base>{};
  }

What does -> signify ? Since the return type is specified as auto, why is it necessary ?

Answer 1

It is referred to as trailing return type, it is simply another way to specify the return type of a function.

One situation where it is useful is returning a function pointer from a function. Here is the "standard" syntax:

void yoyo(){
    std::cout << "yoyo!\n";
}

void(*my_fn())(){
    return yoyo;
}

Versus the more "modern" syntax:

auto my_fn() -> void(*)(){
    return yoyo;
}

Which is much easier to read.

P.S. The trailing return type can be used for function pointer declarations too:

auto yoyo(){ std::cout << "yoyo!\n"; }

auto yoyo_fn() -> void(*)(){ return yoyo; }

auto too_far() -> auto(*)() -> void(*)(){ return yoyo_fn; }

Which is pretty contrived, but is much easier to read than if I had to write in the traditional syntax!