去:从延期返回 [英] Go: returning from defer
问题描述
$ b
func getReport(文件名字符串)我想从函数返回一个错误, (rep报告,错误错误){
rep.data = make(map [string] float64)
推迟func(){
如果r:= recover(); r!= nil {
fmt.Println(在f中恢复,r)
err,_ = r。(错误)
返回nil,err
}
}()
panic(Report format not recognized。)
// getReport函数的其余部分,它可以尝试超出边界访问切片
...
}
我似乎误解了恐慌和延期的概念。任何人都可以启发我吗?
在延迟函数中,您可以更改返回的参数,但不能返回新集。因此,对您所拥有的产品进行简单的更改将会使其发挥作用。
您写的内容还存在另一个问题,即您使用 string
,但在你的类型断言中需要一个错误
。
( Play )
推迟func(){
如果r:= recover(); r!= nil {
fmt.Println(在f中恢复,r)
//找出错误是什么,并设置err
switch x:= r。(type) {
大小写字符串:
err = errors.New(x)
大小写错误:
err = x
默认值:
err = errors.New 未知恐慌)
}
//无效rep
rep = nil
//返回已修改的错误和
}
}()
I want to return an error from a function if it panics (in Go):
func getReport(filename string) (rep report, err error) {
rep.data = make(map[string]float64)
defer func() {
if r := recover(); r != nil {
fmt.Println("Recovered in f", r)
err, _ = r.(error)
return nil, err
}
}()
panic("Report format not recognized.")
// rest of the getReport function, which can try to out-of-bound-access a slice
...
}
I appear to have misunderstood the very concept of panic and defer. Can anybody enlighten me?
In a deferred function you can alter the returned parameters, but you can't return a new set. So a simple change to what you have will make it work.
There is another problem with what you wrote, namely that the you've paniced with a string
but are expecting an error
in your type assertion.
Here is a fix for both of those (Play)
defer func() {
if r := recover(); r != nil {
fmt.Println("Recovered in f", r)
// find out exactly what the error was and set err
switch x := r.(type) {
case string:
err = errors.New(x)
case error:
err = x
default:
err = errors.New("Unknown panic")
}
// invalidate rep
rep = nil
// return the modified err and rep
}
}()
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