英文:
Confusing output on fizzbuzz with switch case statement
问题
以下是使用Go语言中的switch/case和if/else条件语句编写的著名的“Fizz Buzz”程序。问题在于使用switch/case时生成了意外的输出，而if/else（具有相同条件）则正常工作。我知道在Go语言中，switch/case与其他C系语言不同，但是这段代码有什么问题呢？
func main() {
const (
FIZZ = 3
BUZZ = 5
)
// 使用switch/case的部分生成了意外的输出
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("现在使用if/else\n")
// 使用if/else的部分按预期工作
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)
}
}
以上是要翻译的内容。
英文:
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)
}
}
答案1
得分: 6
根据golang的规范，"fallthrough"语句用于在"switch"语句的表达式中将控制转移到下一个case子句的第一条语句。它只能作为该子句中最后一个非空语句使用。
所以问题是："case i % FIZZ == 0"在末尾使用了fallthrough，因此"case i % BUZZ == 0"分支也会被执行，但是条件"i % BUZZ == 0"没有被检查。
因此，要在golang中使用switch实现Fizz Buzz，你需要移除fallthrough，并在顶部添加一个额外的case分支。你可以在play.golang.org上看到，"if-version"更加简洁。
英文:
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.
答案2
得分: 3
你可以使用i%15来实现fizzbuzz。这种方法可以提高性能。一个数字只需要进行一次除法和一次系统调用（sys_write）。而且不需要担心fallthrough的问题。点击这里进行演示。
func main() {
const (
FIZZ = 3
BUZZ = 5
FIZZBUZZ = 15
)
for i := 1; i <= 30; i++ {
switch {
case i % FIZZBUZZ == 0:
fmt.Printf("%d fizzbuzz\n", i)
case i % FIZZ == 0:
fmt.Printf("%d fizz\n", i)
case i % BUZZ == 0:
fmt.Printf("%d buzz\n", i)
default:
fmt.Printf("%d\n", i)
}
}
}
英文:
Your can use i%15 for fizzbuzz. This option gives a gain in performance. One number - one division and one system call (sys_write). And no worries about fallthrough. Play.
<!-- language: go -->
func main() {
const (
FIZZ = 3
BUZZ = 5
FIZZBUZZ = 15
)
for i := 1; i <= 30; i++ {
switch {
case i % FIZZBUZZ == 0:
fmt.Printf("%d fizzbuzz\n", i)
case i % FIZZ == 0:
fmt.Printf("%d fizz\n", i)
case i % BUZZ == 0:
fmt.Printf("%d buzz\n", i)
default:
fmt.Printf("%d\n", i)
}
}
}
答案3
得分: 0
//示例 - 在1和100之间。
for i := 1; i <= 100; i++ {
output := ""
if i%3 == 0 {
output += "Fizz"
}
if i%5 == 0 {
output += "Buzz"
}
if output == "" {
fmt.Println(i)
} else {
fmt.Println(output)
}
}
英文:
//Example - between 1 and 100.
for i := 1; i <= 100; i++ {
output := ""
if i%3 == 0 {
output += "Fizz"
}
if i%5 == 0 {
output += "Buzz"
}
if output == "" {
fmt.Println(i)
} else {
fmt.Println(output)
}
}
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