Xingzhi Pan wrote:
> The first argument to foldr is of type (a -> b -> a), which takes 2
> arguments. But 'step' here is defined as a function taking 3
> arguments. What am I missing here?
You can think of step as a function of two arguments which returns a
function with one argument (although in reality, as any curried function,
'step' is _one_ argument function anyway):
step :: b -> (a -> c) -> (b -> c)
e.g. 'step' could have been defined as such:
step x g = \a -> g (f a x)
to save on lambda 'a' was moved to argument list.
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