英文:
PANIC in default formatting with Go's log package
问题
当使用log.Println在Go中进行登录时，我经常会得到以下错误信息：
2012/05/13 16:45:50 evaluating %v(PANIC=3)
我不确定我做错了什么，我猜想可能是在我的某个Stringer接口实现中，fmt.Println捕获了一个由我自己生成的panic，以避免由于日志记录失败而导致程序崩溃。
我该如何找出发生了什么？为什么会出现这个错误信息？
英文:
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?
答案1
得分: 3
你是对的，String方法中发生了一次panic。但这与log包无关。Println使用了%v，而%v表示运行String方法。在String方法中发生panic会调用catchPanic。在你的输出中，3是panic的值。
英文:
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.
答案2
得分: 1
没有代码进行检查很难说。为了调试它，也许尝试将log.Println("evaluating", foo)替换为log.Printf("evaluating %#v\n", foo)。它的工作方式有点不同：
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)
}
运行结果：
2012/05/13 11:19:49 evaluating %v(PANIC=3)
2012/05/13 11:19:49 evaluating 42
英文:
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
$
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