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POScala Higher Order Function Little Confused
 

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    <p>I was running the below Scala code in Worksheet:</p> <pre><code>package src.com.sudipta.week2.coursera import scala.math.abs import scala.annotation.tailrec object FixedPoint { println("Welcome to the Scala worksheet") //&gt; Welcome to the Scala worksheet val tolerance = 0.0001 //&gt; tolerance : Double = 1.0E-4 def isCloseEnough(x: Double, y: Double): Boolean = { abs((x - y) / x) / x &lt; tolerance } //&gt; isCloseEnough: (x: Double, y: Double)Boolean def fixedPoint(f: Double =&gt; Double)(firstGuess: Double): Double = { @tailrec def iterate(guess: Double): Double = { val next = f(guess) if (isCloseEnough(guess, next)) next else iterate(next) } iterate(firstGuess) } //&gt; fixedPoint: (f: Double =&gt; Double)(firstGuess: Double)Double def myFixedPoint = fixedPoint(x =&gt; 1 + x / 2)(1)//&gt; myFixedPoint: =&gt; Double myFixedPoint //&gt; res0: Double = 1.999755859375 def squareRoot(x: Double) = fixedPoint(y =&gt; (y + x / y) / 2)(1) //&gt; squareRoot: (x: Double)Double squareRoot(2) //&gt; res1: Double = 1.4142135623746899 def calculateAverate(f: Double =&gt; Double)(x: Double) = (x + f(x)) / 2 //&gt; calculateAverate: (f: Double =&gt; Double)(x: Double)Double def myNewSquareRoot(x: Double): Double = fixedPoint(calculateAverate(y =&gt; x / y))(1) //&gt; myNewSquareRoot: (x: Double)Double myNewSquareRoot(2) //&gt; res2: Double = 1.4142135623746899 } </code></pre> <p>What is making me confused are:</p> <ul> <li>Scala worksheet is showing below for my fixedPoint function</li> </ul> <blockquote> <p>fixedPoint: (f: Double => Double)(firstGuess: Double)Double</p> </blockquote> <p>What is this? Is this function type/ function definition or I am missing the term? Basically how I can explain this function in English?</p> <ul> <li>Scala worksheet is showing below for my calculateAverate function</li> </ul> <blockquote> <p>calculateAverate: (f: Double => Double)(x: Double)Double</p> </blockquote> <p>But it looks to me the return type of my function is Double, but I was expecting Double => Double. The reason is I am going to use this with fixedPoint which expects Double => Double like below:</p> <blockquote> <p>def myNewSquareRoot(x: Double): Double = fixedPoint(calculateAverate(y => x / y))(1)</p> </blockquote> <p>Please help me to understand Higher Order Function / Currying more clearly. Thanks in advance.</p>
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