英文:
Does calling a function break recover()?
问题
我正在使用一个从恐慌中恢复的库，它使用的代码简化为以下内容：
func main() {
defer rec()
panic("X")
}
func rec() {
rec2()
}
func rec2() {
fmt.Printf("recovered: %v\n", recover())
}
这段代码的输出是：
recovered: <nil>
panic: X
...更多的恐慌输出...
值得注意的是，recover()返回的是nil而不是错误。这是预期的行为吗？
英文:
I was using a library which recover()s from panics, and it was using code that simplifies to the following:
func main() {
defer rec()
panic("X")
}
func rec() {
rec2()
}
func rec2() {
fmt.Printf("recovered: %v\n", recover())
}
The output of this is:
recovered: <nil>
panic: X
... more panic output ...
Notably, recover() returns nil instead of the error. Is this intended behavior?
答案1
得分: 3
recover 必须由延迟函数直接调用。
根据语言规范：
如果满足以下任一条件，则
recover的返回值为nil：
panic的参数为nil；- 协程没有发生恐慌；
recover不是由延迟函数直接调用的。
英文:
recover must be called directly by a deferred function.
from the language spec:
> The return value of recover is nil if any of the following conditions
> holds:
>
> - panic's argument was nil;
> - the goroutine is not panicking;
> - recover was not called directly by a deferred function.
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