[1] When passing either an object or data to a function, the parameter is stored as a function. e.g. if f=[p]{}, then f(a) and f(5) have p=a and p=5 respectively. If a=5, then the results are identical. The trap is that if a={ret({ret(5)})}, which is a wrapping function, and you call "a" before passing it to "f", then you get the nested function {ret(5)} instead of the wrapping function {ret({ret(5)})}. Having worked with Copper in practice, it's not too hard to spot the error, but it is one of those things that will catch beginners off guard.
[2] https://github.com/chronologicaldot/CupricBridge/blob/master...