[SOLVED] Confusing output on fizzbuzz with switch case statement

Confusing output on fizzbuzz with switch case statement

Here the famous "fizz buzz" program in Go using switch/case and if/else conditionals. The problem is that using switch/case is generating unexpected output while if/else (with same conditions) works fine. I know that switch/case in golang is different than other C-family languages, but what's wrong with this code fragment?

func main() {
const (
    FIZZ = 3
    BUZZ = 5
)

//section with switch/case gives unexpected output
for i := 1; i <= 30; i++ {
    switch {
    case i % FIZZ == 0:
        fmt.Printf("%d fizz\t", i%3)
        fallthrough
    case i % BUZZ == 0:
        fmt.Printf("%d buzz\t", i%5)
    }
    fmt.Printf("\t%d\n", i)
}

fmt.Printf("now towards the if/else\n")

//section with if/else works as expected
for i := 1; i <= 30; i++ {
    if i % FIZZ == 0 {
        fmt.Printf("%d fizz\t", i%3)
    }
    if i % BUZZ == 0 {
        fmt.Printf("%d buzz\t", i%5)
    }
    fmt.Printf("\t%d\n", i)
}

}

Solution

From the golang spec:

Fallthrough statements

A "fallthrough" statement transfers control to the first statement of the next case clause in a expression "switch" statement. It may be used only as the final non-empty statement in such a clause.

So the problem is: "case i % FIZZ == 0" has fallthrough at the end, so "case i % BUZZ == 0" branch is executed too, but the condition "i % BUZZ == 0" is not checked.

So to implement Fizz Buzz in golang using switch you need to remove fallthrough and add one more case branch to the top: play.golang.org. As you can see, "if-version" is more concise.