std::any_cast without needing the type of the original object

Question

Is it possible to use std::any_cast without putting in the first template argument (the type of the object the any is covering)? I tried using any_cast<decltype(typeid(toCast).name())> but it didn't work.

Also tried to store the objects types from the beginning, but that also didn't work because variables can't store types.

Answer 1

One of the fundamentals principles of C++ is that the types of all objects are known at compile time. This is an absolute rule, and there are no exceptions.

The type of the object in question is std::any. It is convertible to some other type only if that type is also known at compile time.

You will note that std::type_info::name() is not a constexpr expression. The returned string is known only at run time. You are attempting to cast something to an object whose type would only be known at run time. C++ does not work this way.

In general, whenever such a situation occurs, nearly all the time the correct solution will involve virtual methods getting invoked on a base class. You will likely need to redesign your classes to use inheritance and virtual methods; using their common base class instead of std::any; then invoking its virtual methods. In some situations std::variant may also work, too.