CODEWITHSUNDEEP
haskell
Ashwin
Ashwin

Reputation: 13517

How do curried functions work in Haskell?

I am learning Haskell. I got to know that any function in Haskell can take only one argument. So, if you see a function max 2 4; it actually is (max 2) 4. What they say is that 1st 2 is applied (as a parameter) to max which returns a functions, that takes 4 as the parameter. What I fail to understand is what happens when 2 is applied to max? What does it mean that it returns a function called (max 2)?

Let me give another example, to make my question more clear. Take this function: multiply x y x = x*y*z. They say it actually is evaluated this way: ((multiply x) y) z. Now I give this input: multiply 2*4*5 How is this evaluated?

multiply 2

returns (multiply 2) and 4 is applied as parameter:

(multiply 2) 4

Now what does this return -- ((multiply 2) 4) or multiply 8? If it multiplies 4 and 2 at this step, how does Haskell know that it has to do that (because the function can multiply only 3 parameters)?

Upvotes: 1

Views: 235

Answers (1)

seven7e
seven7e

Reputation: 818

Just think it mathematically: suppose there is a function taking two variables: f(x, y). Fix x=2 would give you a new function with one variable: g(y)=f(2, y)

If f(x, y) = max(x, y) which gives the maximum of x and y, g(y) = f(2, y) = max(2, y) gives the maximum of 2 and y.

For f(x, y, z) = x * y * z, g(y, z) = f(2, y, z) = 2 * y * z, and h(z) = g(4, z) = f(2, 4, z) = 2 * 4 * z.

Also you can fix x=2 and z=4 to form p(y) = f(2, y, 4). In Haskell it is

\y -> multiply 2 y 4

For the implementation, Haskell would not actually multiply 2 and 4 because it's lazy evaluated. That is, it would not compute a value until it has to.

Upvotes: 2

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