CODEWITHSUNDEEP
haskelllambdasyntax
Gallaoui
Gallaoui

Reputation: 511

Inner bindings in Haskell

I'm new in Haskell, and I don't get it how this code works. Can anyone please explain it to me how the inner binding works in this example:

f = (\x -> ((\x -> x*x) 3)+x)

when I try f 1 I get 10

Upvotes: 1

Views: 266

Answers (2)

Joachim Breitner
Joachim Breitner

Reputation: 25782

You code is quite obfuscated, so let’s clean it up. We start with

f = (\x -> ((\x -> x*x) 3)+x)

and notice that there are two variables that are both called x – how confusing! So let’s write this as

f = (\x -> ((\y -> y*y) 3)+x)

Next, there is this inner lambda expression. We can give it a name and bind it at the same level as f is bound:

square = (\y -> y*y)
f = (\x -> (square 3)+x)

At this point, I would like to remove unnecessary parenthesis:

square = \y -> y*y
f = \x -> square 3 + x

and finally, instead using a lambda abstraction, we can define square and f as functions. Again, this is equivalent code¹:

square y = y*y
f x = square 3 + x

At this point, you can probably make more sense of this code.

¹ In fact, the compiler might optimize the two different when it comes to inlining, which happens only when all manifest arguments are provided, but that is beyond the scope of this question.

Upvotes: 6

ForceBru
ForceBru

Reputation: 44878

You can safely remove the outer parentheses:

f = \x -> ((\x -> x*x) 3) + x

Now, the \x -> stuff construct is an anonymous function of one argument x. The very inner function \x -> x*x squares any given number. (\x -> x*x) 3 calls it, which gives you 3 * 3 == 9.

The outer function calls the inner one first and adds x to the result. So, you get 9 + 1 == 10.

I don't see anything special about binding here. The functions are independent, as well as the variables x.

Upvotes: 1

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