package main
import "fmt"
func main() {
var i int = 10
switch true {
case i < 20:
fmt.Printf("%v is less than 20\n", i)
i = 100
fallthrough
case i < 19:
fmt.Printf("%v is less than 19\n", i)
fallthrough
case i < 18:
fmt.Printf("%v is less than 18\n", i)
fallthrough
case i > 50:
fmt.Printf("%v is greater than 50\n", i)
fallthrough
case i < 19:
fmt.Printf("%v is less than 19\n", i)
fallthrough
case i == 100:
fmt.Printf("%v is equal to 100\n", i)
fallthrough
case i < 17:
fmt.Printf("%v is less than 17\n", i)
}
}
Output:
10 is less than 20
100 is less than 19
100 is less than 18
100 is greater than 50
100 is less than 19
100 is equal to 100
100 is less than 17
Is this expected behavior?
The fallthrough statement transfers control to the first statement of the next case block.
The fallthrough statement does not mean to continue evaluating the expression of the next case, but to unconditionally start executing the next case block.
Quoting from fallthrough statement doc:
A "fallthrough" statement transfers control to the first statement of the next case clause in an expression "switch" statement.
Quoting from switch statement doc:
In a case or default clause, the last non-empty statement may be a (possibly labeled) "fallthrough" statement to indicate that control should flow from the end of this clause to the first statement of the next clause. Otherwise control flows to the end of the "switch" statement.
Yes it is, as pointed by icza.
If you don't want to fall under every case block after first, remove fallthrough lines (it's like you wouldn't have put a break line at the end of each C/C++ case block.
And, as you expect in comments, the evaluation is done when switch() is reached, after that it doesn't matter if you change i value, it won't be evaluated again on each case block.