Take A `Step` With Kotlin

Jan 21, 2020

This blog post is inspired by Anton Keks's amazing Kotlin Puzzlers talk from KotlinConf '19. The blog post outlines the steps in my thought process while exploring and demystifying the Puzzlers.

So, what would this code snippet output?

fun main() {
    for(i in 1..10 step 2 step 3) print("$i ")
}

Let's understand this, step by step. 🤭

What would the output be if we did only a step 2 in the loop, like this:

for(i in 1..10 step 2) print("$i ")

Unsurprisingly, this would output 1 3 5 7 9 - in the range 1..10, we increment by steps of 2.

So now coming back to our original question, what happens when we do step 2 step 3?

To understand this, we first need to clearly understand what exactly happens behind the scenes when we do 1..10 step 2. Let's just print this out -

fun main() {
    println(1..10 step 2) //Output: 1..9 step 2
}

Okay, that's a funny output, where did that come from? Let's dive deeper.

Let's navigate to Koltin source codes to inspect behavior.

Navigating to step implementation (Cmd+B in IntelliJ), we see this is how it looks:

public infix fun IntProgression.step(step: Int): IntProgression {
    checkStepIsPositive(step > 0, step)
    return IntProgression.fromClosedRange(first, last, if (this.step > 0) step else -step)
}

Glaring at us from this code snippet is IntProgression.

Navigating to IntProgression implementation (Cmd+B in IntelliJ), we see that this is a class with start, endInclusive and step -

public open class IntProgression
    internal constructor
        (
                start: Int,
                endInclusive: Int,
                step: Int
        )
    ...

Scrolling down a bit, we see its toString() function:

override fun toString(): String = if (step > 0) "$first..$last step $step" else "$first downTo $last step ${-step}"

There we go! This method is what leads to printing the not-so-funny-anymore 1..9 step 2 as output - string representation of IntProgression with values start -> 1, endInclusive -> 9 (computed in fromClosedRange -> getProgressionLastElement) and step 2.

Thus, to summarize 1..10 step 2 returns an IntProgression!

Now that we've understood what happens behind the scenes of step, we are right on track to reason out the solution for the original question:

for(i in 1..10 step 2 step 3) print("$i ")

1..10 step 2 returns an IntProgression in the range 1..9.

Thus, we now actually boil down to a no puzzler and basically evaluate -

for(i in 1..9 step 3) print("$i ")

And the output is surely,

1 4 7

Tada! 🎉

Take A Step With Kotlin

Take A `Step` With Kotlin