[chancho]

You've mixed together a lot of stuff here. your Foo data type is recursive. I can achieve something similar in C++ without infinite template recursion, but I have to use pointers to do it. (Your Haskell has pointers too, they're just hidden from you.) Also there is nothing 'runtime' about the example you've got here. I can do the same thing with a tagged-union type in C/C++ and a switch statement, which is what your Haskell compiles down to. Maybe you were thinking of existential types? http://en.wikibooks.org/wiki/Haskell/Existentially_quantifie... . That's something C++ can't do.

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Edit: nm I see what you did there with the Foo a = ... | Blah (Foo [a]). That would probably break C++.