Because applying f to the list is not the same thing is applying bind to the list and f.
Bob On 26 Feb 2011, at 20:17, michael rice wrote: > Why? Shouldn't this work for any type a? > > Michael > > ========== > > f :: [a] -> [a] > f l = do x <- l > return x > > ========== > > *Main> :r > [1 of 1] Compiling Main ( test.hs, interpreted ) > Ok, modules loaded: Main. > *Main> f "abcde" > "abcde" > *Main> f [1,2,3,4,5] > [1,2,3,4,5] > *Main> "abcde" >>= f > > <interactive>:1:12: > Couldn't match expected type `Char' against inferred type `m b' > In the second argument of `(>>=)', namely `f' > In the expression: "abcde" >>= f > In the definition of `it': it = "abcde" >>= f > *Main> > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe
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