What happens if the expression in a match arm returns false?

Question

This would be a nice fizzbuzz in Rust I think:

match (i % 3, i % 5) {
    (0, 0) => println!("FizzBuzz"),
    (0, _) => println!("Fizz"),
    (_, 0) => println!("Buzz"),
    _ => println!("{}", i),
}   

It could also be stated this way:

match i {
    i if i % 3 == 0 && i % 5 == 0 => println!("FizzBuzz"),
    i if i % 3 == 0 => println!("Fizz"),
    i if i % 5 == 0 => println!("Buzz"),
    _ => println!("{}", i),
}

Now that confused me.

i if i % 3 == 0

is an expression, right? So, this evaluates to i when the condition is true. But what happens if it is false?

Answer 1

Now that confused me.

i if i % 3 == 0

is an expression, right?

No, it is not. It is a pattern (i) with a guard (i % 3 == 0). Maybe you got confused because you used the same bind name. Consider this modified example:

match i {
    x if x % 3 == 0 && x % 5 == 0 => println!("FizzBuzz"),
    x if x % 3 => println!("Fizz"),
    x if x % 5 => println!("Buzz"),
    _ => println!("{}", x),
}

You can read the match expression like this

• If i matches the pattern x (it always match and i value is moved (copied) to x) and x % 3 == 0 and x % 5 == 0 then println!("FizzBuzz"); else
• if i matches the pattern x and x % 3 == 0 then println!("Fizz"); else
• if i matches the pattern x and x % 5 == 0 then println!("Buzz"); else
• println!("{}", x)

Answer 2

If it is false, the match arm is simply not called. These are called guards in the match statement. You can read about them in the book.

They are similar to an if-else but not in the case that there is an else block. They just add more filtering to the match blocks.