Hi there, The problem is in these lines:
| i < i' = [] | i == i' = [] | i > i' = []
Comparisons are defined by the library, not the language. There is no way for the compiler to know they're mutually exclusive, short of solving the halting problem. In fact, it is possible to write this: instance Ord MyType where _ < _ = False _ == _ = False _ > _ = False and your f will fail at run time. The solution is to either replace your last comparison with "otherwise" (since it must be True anyway), or use `compare`: f i (c:cs) ((i',t):os) = case i `compare` i' of LT -> ... EQ -> ... GT -> ... Since these cases *are* mutually exclusive, it will pass the warning.
main :: IO () main = do print $ f 0 [] []
without the "otherwise" clause (commented) I get the warning
example.hs:2:1: Warning: Pattern match(es) are non-exhaustive In an equation for `f': Patterns not matched: _ (_ : _) ((_, _) : _)
Could someone help me uncover which cases that are not… erm… covered by the above patterns?
Many thanks, Semen
PS This definition, on the other hand,
f _ cs [] = cs f _ [] _ = [] f i (c:cs) ((i',t):os) = []
is exhaustive…
-- Семен Тригубенко http://trygub.com
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