Hi Paul,
On Dec 18, 2007 5:18 PM, Paul Hudak
If the semantics of a language says that a function f is equivalent to a function g, but there is a function h such that h(f) is not equivalent to h(g), then h cannot be a function.
Sure.
Therefore that language cannot be a (purely) functional language.
That is the pure and simple reason why functions are not Showable in Haskell.
Not so fast :-) Caveat one, there may be useful ways to for functions to implement Show that don't conflict with extensionality (i.e., the property that two functions are equal if they yield the same results for all inputs). Caveat two, we generally assume extensionality when reasoning about Haskell, but it's entirely possible to give a semantics for Haskell that doesn't assume extensionality. IMHO, a good answer to the question why functions aren't showable in Haskell needs to explain why we prefer our semantics to be extensional, not say that by god-given fiat, Haskell is extensional, so we can't show functions. - Benja