Simple Fizz Buzz Program

The FizzBuzz program is a common coding challenge used to assess basic programming skills. The program iterates through a sequence of numbers, typically from 1 to 100. For each number, it checks for divisibility by 3 and 5. If a number is divisible by 3, it prints "Fizz". If it's divisible by 5, it prints "Buzz". If it's divisible by both 3 and 5, it prints "FizzBuzz". If none of the conditions are met, it prints the number itself.

Now, let's try to write the FizzBuzz program in redy.

fn main() {
  for i: 1..=100 {
    if i % 15 == 0 {
      println#("FizzBuzz")
    } else if i % 3 == 0 {
      println#("Fizz")
    } else if i % 5 == 0 {
      println#("Buzz")
    } else {
      println#("{}", i)
    }
  }
}

If you run this code, you will get the FizzBuzz output from 1 to 100. Let's understand what's going on here.

The first line for i: 1..=100 { ... } creates a loop that iterates a number i from 1 to 100 (inclusive). The code inside the curly braces { ... } will be executed for each number.

Inside the loop, we use if...else if...else statements, which are used to execute different code blocks based on conditions.

• i % 15 == 0: This condition checks if the number i is perfectly divisible by 15. The % operator calculates the remainder of a division. If the remainder is 0, it means i is a multiple of 15.
• i % 3 == 0 and i % 5 == 0: These conditions check for divisibility by 3 and 5, respectively.
• else { println#("{}", i) }: If none of the above conditions are true, the program executes the code in this block, simply printing the number itself.

Notice that the condition i % 15 == 0 comes first. This is because a number divisible by 15 is also divisible by 3 and 5. If we checked for i % 3 == 0 first, it would print "Fizz" and then move on, never reaching the "FizzBuzz" condition.