In Haskell, you can't ever check that a type doesn't implement a type class.

In Golang, a type can only implement an interface if the implementation is defined in the same module as the type.

In C++, in typical C++ style, it's the wild west and the compiler doesn't put guard rails on, and does what you would expect it to do if you think about how the compiler works, which probably isn't what you want.

I don't know what Rust does.