[SamReidHughes_]

C++ requires that all overloaded versions exist at compile time; Haskell allows them to exist at runtime. For example, in Haskell you can make a datatype like

    data Foo a = Blah a | Bar (Foo [a])

and you'll have no trouble making a function "showFoo :: Show a => Foo a -> String" using straightforward programming. If you try to do that in C++ you'll get infinite template recursion.

Also, C++ templates don't restrict you to match the right type signatures, the way Haskell typeclasses do.

But other than that, they are pretty much the same kind of thing and are used in the same way.

[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++.