If your function is currently taking a parameter by-value, and you replace it with std::optional<T>, you're still taking it by value and doing the exact same amount of copies. No (new) issues from optional.

You only run into trouble if you do something like change `void foo(const T&)` to 'void foo(std::optional<T>)`. But in such a change, it's "clear" (insofar as C++ is ever clear about anything) you're replacing a reference parameter with a value parameter.

I don't think this is surprising std::optional behavior?

> The (typical C++ stdlib) problem here is that things that look deceptively simple on the surface may result in a surprising amount of activity under the hood

It's literally exactly what it looks like. The C++ stdlib didn't complicate or obfuscate anything.