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

this nice fizzbuzz in rust think:

match (i % 3, % 5) {     (0, 0) => println!("fizzbuzz"),     (0, _) => println!("fizz"),     (_, 0) => println!("buzz"),     _ => println!("{}", i), }    

it stated way:

match {     if % 3 == 0 && % 5 == 0 => println!("fizzbuzz"),     if % 3 == 0 => println!("fizz"),     if % 5 == 0 => println!("buzz"),     _ => println!("{}", i), } 

now confused me.

i if % 3 == 0 

is expression, right? so, evaluates i when condition true. happens if false?

now confused me.

i if % 3 == 0 

is expression, right?

no, not. pattern (i) guard (i % 3 == 0). maybe got confused because used same bind name. consider modified example:

match {     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 match expression this

• if i matches pattern x (it match , i value moved (copied) x) , x % 3 == 0 , x % 5 == 0 println!("fizzbuzz"); else
• if i matches pattern x , x % 3 == 0 println!("fizz"); else
• if i matches pattern x , x % 5 == 0 println!("buzz"); else
• println!("{}", x)