Doesn't constraining the "auto" in C++ defeat the purpose of it?

Question

In C++20, we are now able to constrain the auto keyword to only be of a specific type. So if I had some code that looked like the following without any constraints:

auto something(){
  return 1;
}

int main(){
  const auto x = something();
  return x;
}

The variable x here is deduced to be an int. However, with the introduction of C++20, we can now constrain the auto to be a certain type like this:

std::integral auto something(){
  return 0;
}

int main(){
  const auto x = something();
  return x;
}

Doesn't this defeat the purpose of auto here? If I really need a std::integral datatype, couldn't I just omit the auto completely? Am I misunderstanding the use of auto completely?

Answer 1

A constraint on the deduced auto type doesn't mean it needs to be a specific type, it means it needs to be one of a set of types that satisfy the constraint. Note that a constraint and a type are not the same thing, and they're not interchangeable.

e.g. a concept like std::integral constrains the deduced type to be an integral type, such as int or long, but not float, or std::string.

If I really need a std::integral datatype, couldn't I just omit the auto completely?

In principle, I suppose you could, but this would at the minimum lead to parsing difficulties. e.g. in a declaration like

foo f = // ...

is foo a type, or a constraint on the type?

Whereas in the current syntax, we have

foo auto f = // ...

and there's no doubt that foo is a constraint on the type of f.