# Solution Review: Using a Curried Function

In the following lesson, we will go over the solution of the challenge: Using a Curried Function.

We'll cover the following

In this challenge, you had to create a non-recursive factorial function fact in terms of a curried function product which calculates the product of the values of a function for the points on a given interval.

## Solution

A skeleton of the function was already provided for you. Let’s look it over.

def product(f: Int => Int)(a: Int, b: Int): Int ={
if(a > b) 1
else f(a) * product(f)(a+1,b)
}

def fact(n: Int) = {

}


product is a recursive curried function and has 2 parameter lists.

1. The first parameter list contains a single parameter; a function which has a single parameter of type Int and returns an integer.
2. The second parameter list contains two parameters a and b, both of type Int. a represented the minimum bound of the interval and b represented the maximum bound.

fact has a single parameter n of type Int. n is the integer in question whose factorial is to be computed.

To write the function body of fact, you needed to call product and pass it an anonymous function that acts as an identifier and returns the integer as is.

x => x


The second argument to be passed to product was the interval a-b. Since a factorial of a number n is simply the product of 1 and all consecutive numbers until n, our range was simply:

(1, n)


You can find the complete solution below:

You were required to write the code on line 7.