可以在defer函数内部惊慌吗，尤其是当它已经惊慌时？

Question

func sub(){
    defer func (){
        panic(2)
    }()
    panic(1)
}

func main(){
    defer func(){
        x:=recover()
        println(x.(int));
    }()
    sub()
}

I tried this code and it seems the first panic panic(1) is simply "overwritten" by the second panic panic(2).

But is it okay to do that? Or call a Golang function that might panic inside defer function?

(In C++ it's almost never acceptable to throw exception out of a destructor. It terminates program if stack is already unwinding. I wonder if panicking in a similar manner could be bad in Golang.)

Answer 1

Yes, it's okay. Panicking from a deferred function is not really a new, special state, it just means that the panicking sequence will not stop.

Your example code also proves that it's okay, and even a panic() called from a deferred function can be stopped by an "upper" level call to recover().

Spec: Handling panics:

Suppose a function G defers a function D that calls recover and a panic occurs in a function on the same goroutine in which G is executing. When the running of deferred functions reaches D, the return value of D's call to recover will be the value passed to the call of panic. If D returns normally, without starting a new panic, the panicking sequence stops. In that case, the state of functions called between G and the call to panic is discarded, and normal execution resumes.

One thing to note here is that even if you call panic() in a deferred function, still all the other deferred functions will run. Also a panic() without recover() from a deferred function will rather "wrap" the existing panic and not "overwrite" it (although it's true that a recover() call will only give you back the value passed to the last panic() call).

See this example:

func main() {
    defer func() {
        fmt.Println("Checkpoint 1")
        panic(1)
    }()
    defer func() {
        fmt.Println("Checkpoint 2")
        panic(2)
    }()
    panic(999)
}

Output (try it on the Go Playground):

Checkpoint 2
Checkpoint 1
panic: 999
    panic: 2
    panic: 1

goroutine 1 [running]:
panic(0xfed00, 0x1040e140)
    /usr/local/go/src/runtime/panic.go:500 +0x720
main.main.func1()
    /tmp/sandbox284410661/main.go:8 +0x120
panic(0xfed00, 0x1040e0fc)
    /usr/local/go/src/runtime/panic.go:458 +0x8a0
main.main.func2()
    /tmp/sandbox284410661/main.go:12 +0x120
panic(0xfed00, 0x1040e0f8)
    /usr/local/go/src/runtime/panic.go:458 +0x8a0
main.main()
    /tmp/sandbox284410661/main.go:14 +0xc0

Even though all deferred functions call panic(), all deferred functions get executed, and the final panic sequence printed contains values passed to all panic() calls.

If you call recover() in the deferred functions, you also get this "recovered" state or info in the final printout:

defer func() {
    recover()
    fmt.Println("Checkpoint 1")
    panic(1)
}()
defer func() {
    recover()
    fmt.Println("Checkpoint 2")
    panic(2)
}()

Output (try it on the Go Playground):

Checkpoint 2
Checkpoint 1
panic: 999 [recovered]
    panic: 2 [recovered]
    panic: 1
...