Why would it need to copy anything? What's wrong with just evaluating the default argument expression every time the function is called with a default argument rather than once at definition-time? That's how it works in other languages.
That's solved easily enough by evaluating the expression in the scope where it was defined (again that's what other languages do).
> Also, I'm rather sure that's how it works in C++
Default arguments in C++ work as I described: If you define `void f(int arg = g(x)) {}`, then calling `f()` n times will call `g` n times as well (using the `x` that was in scope when `f` was defined) - and if you never call `f`, `g` isn't called either.
An example demonstrating this: https://ideone.com/vs26Oq