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LESSON 116 OF 348 · WHILE LOOPS & FOR LOOPS

Printing Numbers with a While Loop in Kotlin

Lesson Read the full explanation

We can achieve many different outcomes with while loops.

For example, we can print print all the numbers from 0 through 4:

var i = 0
while (i < 5) {
	println(i)
	i++
}

The output of this program is:

0
1
2
3
4

And this is what the computer does behind the scenes during this loop:

0. Creates and initializes a variable i = 0

1. First iteration:
	a. Is i < 5 true? <=> Is 0 < 5 true? Yes. 
	b. println(i) => Output: 0
	c. i++ => i = 1
	
1. Second iteration:
	a. Is i < 5 true? <=> Is 1 < 5 true? Yes. 
	b. println(i) => Output: 1
	c. i++ => i = 2
	
2. Third iteration:
	a. Is i < 5 true? <=> Is 2 < 5 true? Yes. 
	b. println(i) => Output: 2
	c. i++ => i = 3
	
3. Forth iteration:
	a. Is i < 5 true? <=> Is 3 < 5 true? Yes. 
	b. println(i) => Output: 3
	c. i++ => i = 4
	
4. Fifth iteration:
	a. Is i < 5 true? <=> Is 4 < 5 true? Yes. 
	b. println(i) => Output: 4
	c. i++ => i = 5

5. Sixth iteration:
	a. Is i < 5 true? <=> Is 5 < 5 true? No.
	b. Exit the loop.

Assignment
Let's print all numbers from 3 to 10 using a while loop.

Hint
Look at the examples above if you get stuck.

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    KOTLIN · WHILE LOOPS & FOR LOOPS

    Printing Numbers in Kotlin, explained

    Counting with a while loop

    A while loop can produce a sequence of numbers if you let a variable hold the current number, print it, and move it forward on every pass. This program prints the numbers from 0 through 4:

    var i = 0
while (i < 5) {
    println(i)
    i++
}

    The output is five lines:

    0
1
2
3
4

    Tracing it shows the pattern. Before the loop, i is 0. Each time the condition i < 5 is checked and found true, the body prints i and then adds 1. After printing 4, i becomes 5, the check 5 < 5 is false, and the loop ends. The value 5 is never printed, because the check happens before the body.

    Initialization and condition

    Two choices decide which numbers appear:

    • The initial value is the first number printed. Starting i at 0 prints 0 first, and starting it at 3 prints 3 first.
    • The condition decides the last number. With i < 5 the last printed value is 4. With i <= 5 it would be 5. The difference between < and <= is the usual source of off-by-one errors.

    Tracing the loop

    For the exercise below, the trace begins with counter at 3. Each pass prints the current value and then adds 1, so the printed values are 3, 4, 5 and so on. When the counter reaches 10, the condition 10 <= 10 is still true, so 10 is printed too. Only when the counter becomes 11 does the test fail. Reading a loop this way, one value at a time, is the most reliable way to find mistakes in it.

    The exercise

    The task is to print every number from 3 to 10, one per line. It has four steps.

    1. Declare the counter with var counter = 3. The first number to print is 3, so that is where it starts. It is a var because the loop will change it, and a val can't be reassigned.
    2. Write the loop with the condition counter <= 10. The task includes 10, so the condition must stay true when counter is 10. With < 10 the last number printed would be 9.
    3. Print the counter with println(counter). The function ends the line after the value, so each number is on its own line.
    4. Increase the counter with counter++. Without this line, counter stays 3, the condition is always true, and the loop never ends.
    var counter = 3
while (counter <= 10) {
    println(counter)
    counter++
}

    The output is the numbers 3, 4, 5, 6, 7, 8, 9 and 10, each on its own line. The loop body runs eight times, because 10 - 3 + 1 = 8. After the loop, counter holds 11.

    Checking your work

    Check a loop by looking at its first and last printed values. The first is the starting value, 3. The last is the largest value that still passes the condition: 10 passes 10 <= 10, while the next value, 11, fails it and ends the loop. If the output stops at 9 or runs to 11, the condition is wrong.

    Gotchas

    • Off by one. To include the last number, use <= with that number, or < with the number after it. Both counter <= 10 and counter < 11 work here.
    • Update before print. If counter++ comes first in the body, the first number printed is 4 and the last is 11.
    • Printing text instead of the value. println("counter") prints the word, not the number.
    • Using val. counter++ on a val is the compile error 'val' cannot be reassigned.

    Key takeaways

    • The initial value of the counter is the first number the loop handles.
    • The condition is checked before each pass, so a number that fails it is never printed.
    • Use <= when the final number is included.
    • Update the counter inside the body, or the loop runs forever.
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