The problem
The aim of this problem is to jot down a higher-order perform returning a brand new perform that iterates on a specified perform a given variety of occasions. This new perform takes in an argument as a seed to start out the computation.
As an illustration, take into account the perform getDouble. When run twice on worth 3, yields 12 as proven under.
getDouble(3) => 6
getDouble(6) => 12
Allow us to identify the brand new perform createIterator and we should always be capable to receive the identical outcome utilizing createIterator as proven under:
var doubleIterator = createIterator(getDouble, 2); // Runs *getDouble* twice
doubleIterator(3) => 12
For the sake of simplicity, all perform inputs to createIterator would perform returning a small quantity and the variety of iterations would at all times be integers.
The answer in Golang
Possibility 1:
package deal resolution
func CreateIterator(fn func(int) int,n int) func(int) int {
return func (i int) int {
for j := 0; j < n; j++ { i = fn(i) }
return i
}
}
Possibility 2:
package deal resolution
func CreateIterator(fn func(int) int,n int) func(int) int {
if n == 0 {
return func(x int) int { return x } // id
} else {
return func(x int) int { return CreateIterator(fn, n-1)(fn(x)) }
}
}
Possibility 3:
package deal resolution
func CreateIterator(fn func(int) int,n int) func(int) int {
if n < 1 {
return nil
}
var retFunc = func(r int) int {
for i := n ; i>0 ; i-- {
r = fn(r)
}
return r
}
return retFunc
}
Check instances to validate our resolution
package deal solution_test
import (
. "github.com/onsi/ginkgo"
. "github.com/onsi/gomega"
)
var _ = Describe("Iterator for 'getDouble' perform",func() {
var getDouble = func(n int) int {return 2*n}
It("Working the iterator as soon as",func() {
doubleIterator := CreateIterator(getDouble,1)
Anticipate(doubleIterator(3)).To(Equal(6))
Anticipate(doubleIterator(5)).To(Equal(10))
})
It("Working the iterator twice",func() {
getQuadruple := CreateIterator(getDouble,2)
Anticipate(getQuadruple(2)).To(Equal(8))
Anticipate(getQuadruple(5)).To(Equal(20))
})
})
