Il 16 gennaio 2021 alle 16:10 Lawrence Bottorff ha scritto:
I have this
myLength1 = foldl (\n _ -> n + 1) 0
and this
myLength2 = foldr (\_ n -> n + 1) 0
I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to the independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well
:t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
foldl and foldr have slightly different signatures, λ> :t +d foldl foldl :: (b -> a -> b) -> b -> [a] -> b λ> :t +d foldr foldr :: (a -> b -> b) -> b -> [a] -> b (Notice `b -> a -> b` vs. `a -> b -> b`), hence the lambdas have a different non-matched parameter. Does this answer your question? —F