Hi gophers,
I was playing around with a puzzle trying to break the sync package and
found something cool. Can you think of a definition for f that causes
once.Do to execute the argument more than once?
package main
import (
"sync"
)
func main() {
var once sync.Once
var f = // ...
once.Do(f)
}
HIGHLIGHT BELOW FOR ANSWER
package main
import (
"fmt"
"sync"
)
func main() {
var once sync.Once
var f func()
times := 9
f = func() {
if times == 0 {
return
}
times--
fmt.Println("Called")
oldonce := once
*&once = sync.Once{}
once.Do(f)
once = oldonce
}
once.Do(f)
}
HIGHLIGHT ABOVE FOR ANSWER
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