Couldn't skip type annotations as that raises ambiguous type error.
Explanation:
Function foldl applies a given function to all container elements going from inside out. When applied to list, it effectively traverses list in a reverse order.
Binary operator foldl is applied to flip (:) and [].
foldl has type signature Foldable t => (b -> a -> b) -> b -> t a -> b, which can be rewritten as ([a] -> a -> [a]) -> [a] -> [a] -> [a].
Binary operator (:) appends a value (first argument) to a list (second argument). It's type signature is a -> [a] -> [a]. It almost looks like the required signature from the 3. point.
Use the flip function in order to switch around (:) first and second argument. This results in a function flip (:) with type signature of [a] -> a -> [a].
Now when this flip (:) is partially applied to foldl, it gives a function foldl (flip (:)) with a type signature [a] -> [a] -> [a]. Giving foldl a initial value of [], gives the final form foldl (flip (:)) [] with a type signature [a] -> [a].
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Haskell solution in exactly 28 characters:
Longer version of the same, but more readable:
Couldn't skip type annotations as that raises ambiguous type error.
Explanation:
foldlapplies a given function to all container elements going from inside out. When applied to list, it effectively traverses list in a reverse order.foldlis applied toflip (:)and[].foldlhas type signatureFoldable t => (b -> a -> b) -> b -> t a -> b, which can be rewritten as([a] -> a -> [a]) -> [a] -> [a] -> [a].(:)appends a value (first argument) to a list (second argument). It's type signature isa -> [a] -> [a]. It almost looks like the required signature from the 3. point.flipfunction in order to switch around(:)first and second argument. This results in a functionflip (:)with type signature of[a] -> a -> [a].flip (:)is partially applied tofoldl, it gives a functionfoldl (flip (:))with a type signature[a] -> [a] -> [a]. Givingfoldla initial value of[], gives the final formfoldl (flip (:)) []with a type signature[a] -> [a].