Understanding the second input to repeatFor

Step 1

Type in the following code and run it:

clearOutput()
def printBoth(n1: Int, n2: Int) {
    println(n1)
    println(n2)
}

printBoth(10, 12)

Q1a. Is printBoth a command or a function?

Q1b. On which line is printBoth defined?

Q1c. On which line is printBoth used?

Q1d. How many inputs does printBoth take?

Step 2

Type in the following code and run it:

clearOutput()
def printBothV2(n1: Int)(n2: Int) {
    println(n1)
    println(n2)
}

printBothV2(10)(12)

Q2a. How is the definition of printBothV2 different from the definition of printBoth?

Q2b. How is the usage of printBothV2 different from the usage of printBoth?

Step 3

Type in the following code and run it:

clearOutput()
def repeatForBoth(n1: Int, n2: Int)(cmd: Int => Unit) {
    cmd(n1)
    cmd(n2)
}

def printInt(n: Int) {
    println(n)
}

repeatForBoth(10, 12)(printInt)

repeatForBoth(10, 12) { n =>
    println(n)
}

Q3a. What are the inputs to repeatForBoth?

Notice that the cmd input to repeatForBoth is itself a command. It’s type is Int => Unit - which means - a function that takes an Int and returns Unit (if you remember, anything that returns Unit is a command).

Q3b. How does repeatForBoth make use of its inputs?

Q3c. How are the two usages of repeatForBoth different? Do you see how the first call to repeatForBoth uses a named command (defined earlier), while the second call uses an unnamed command (defined right at the point of usage). Notice how the input and body of the anonymous command are separated by a =>.

Step 4

Type in the following code and run it:

clearOutput()

val ab = ArrayBuffer(2, 9, 3)

def printInt(n: Int) {
    println(n)
}

repeatFor(ab)(printInt)

repeatFor(ab) { n =>
    println(n)
}

Q4a. Describe what the above code is doing.

Exercise - Create an ArrayBuffer with the first 5 prime numbers as elements. Then use repeatFor to go through (or iterate over) the elements and print them.