How does foldr work?

haskell combinators fold
dinsim picture dinsim · Nov 18, 2009 · Viewed 67.5k times · Source

Can anybody explain how does foldr work?

Take these examples:

Prelude> foldr (-) 54 [10, 11]
53
Prelude> foldr (\x y -> (x+y)/2) 54 [12, 4, 10, 6]
12.0

I am confused about these executions. Any suggestions?

Answer

Tirpen picture Tirpen · Nov 19, 2009

The easiest way to understand foldr is to rewrite the list you're folding over without the sugar.

[1,2,3,4,5] => 1:(2:(3:(4:(5:[]))))

now what foldr f x does is that it replaces each : with f in infix form and [] with x and evaluates the result.

For example:

sum [1,2,3] = foldr (+) 0 [1,2,3]

[1,2,3] === 1:(2:(3:[]))

so

sum [1,2,3] === 1+(2+(3+0)) = 6

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