Reputation: 118710
When logging in Go with log.Println
, I frequently get
2012/05/13 16:45:50 evaluating %v(PANIC=3)
I'm not sure how to determine what I've done wrong, I assume that somewhere fmt.Println
has caught a panic generated by one of my own Stringer
interface implementations, so as not to crash my program due to logging failure.
How do I work out what's going on? Why am I getting this erroneous message?
Upvotes: 1
Views: 1501
Reputation: 28126
You are right, there is a panic in a String
method. But it has nothing to do with the log
package. Println
uses %v
, and %v
means running String
method. Having a panic in the String
method invokes catchPanic
. Here in your output 3
is the value of your panic.
Upvotes: 3
Reputation: 91429
W/o the code to inspect it's hard to say. To debug it, perhaps try replacing log.Println("evaluating", foo)
with log.Printf("evaluating %#v\n", foo)
. It works a bit differently:
package main
import "log"
type t int
func (t) String() string {
panic(3)
}
func main() {
var v t = 42
log.Println("evaluating", v)
log.Printf("evaluating %#v\n", v)
}
$ go run main.go
2012/05/13 11:19:49 evaluating %v(PANIC=3)
2012/05/13 11:19:49 evaluating 42
$
Upvotes: 1