Haskell用匿名函数折叠

Question

I have a problem with one of the Haskell basics: Fold + anonymous functions

I'm developing a bin2dec program with foldl .
The solution looks like this:

bin2dec :: String -> Int
bin2dec = foldl (\x y -> if y=='1' then x*2 + 1 else x*2) 0

I understand the basic idea of foldl / foldr but I can't understand what the parameters xy stands for.

Answer 1

See the type of foldl

foldl :: (a -> b -> a) -> a -> [b] -> a

Consider foldl fz list

so foldl basically works incrementally on the list (or anything foldable), taking 1 element from the left and applying fz element to get the new element to be used for the next step while folding over the rest of the elements. Basically a trivial definition of foldl might help understanding it.

 foldl f z []     = z
 foldl f z (x:xs) = foldl f (f z x) xs

The diagram from Haskell wiki might help building a better intuition.

Consider your function f = (\\xy -> if y=='1' then x*2 + 1 else x*2) and try to write the trace for foldl f 0 "11" . Here "11" is same as ['1','1']

  foldl f 0 ['1','1'] 
= foldl f (f 0 '1') ['1']

Now f is a function which takes 2 arguments, first a integer and second a character and returns a integer. So In this case x=0 and y='1' , so fxy = 0*2 + 1 = 1

= foldl f 1 ['1']
= foldl f (f 1 '1') []

Now again applying f 1 '1' . Here x=1 and y='1' so fxy = 1*2 + 1 = 3 .

= foldl f 3 [] 

Using the first definition of foldl for empty list.

= 3 

Which is the decimal representation of "11".

Answer 2

Use the types! You can type :t in GHCi followed by any function or value to see its type. Here's what happens if we ask the for the type of foldl

Prelude> :t foldl
foldl :: (a -> b -> a) -> a -> [b] -> a

The input list is of type [b] , so it's a list of b s. The output type is a , which is what we're going to produce. You also have to supply an initial value for the fold, also of type a . The function is of type

a -> b -> a

The first parameter ( a ) is the value of the fold computed so far. The second parameter ( b ) is the next element of the list. So in your example

\x y -> if y == '1' then x * 2 + 1 else x * 2

the parameter x is the binary number you've computed so far, and y is the next character in the list (either a '1' or a '0' ).