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?
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.
Related
I was referred to this question: Program recovered from panic does not exit as expected
It works fine but it relies on knowing where the panic occurs in order to place the deferred function.
My code is as follows.
package main
import "fmt"
func main() {
defer recoverPanic()
f1()
f2()
f3()
}
func f1() {
fmt.Println("f1")
}
func f2() {
defer f3() //<--- don't want to defer f3 here because I might not know f2 will panic, panic could occuer elsewhere
fmt.Println("f2")
panic("f2")
}
func f3() {
fmt.Println("f3")
}
func recoverPanic() {
if r := recover(); r != nil {
fmt.Printf("Cause of panic ==>> %q\n", r)
}
}
Having the deferred function call f3() in the panicking function works, output below.
f1
f2
f3
Cause of panic ==>> "f2"
What if you have an application where you don't know where a panic occurs, do I need to put a defer in every function that might panic?
Commenting out the defer f3() gives me the following output.
f1
f2
Cause of panic ==>> "f2"
f3 never runs.
My question is how to continue execution of the program without having a deferred function call in every function that might panic?
You can't resume function execution after a panic. Panic is used when the current line of execution cannot continue correctly. Arbitrarily resuming execution after a panic (if it were possible) is begging immediately for another panic, because the state is already incorrect and just blazing ahead won't fix that.
For example, let's say a function panics when it tries to read out of bounds on a slice. How can it continue? What would it even mean to continue? Should it just read the out of bounds memory location and get garbage data? Continue with a zero value? Take a different value from the slice?
You must handle error cases; either by explicitly recovering, or preemptively checking / correcting conditions that will result in panic. At least in the standard library, functions that may spur a panic will say so in their documentation with an explanation of which conditions will result in panic.
If you commonly need to safely call void functions and recover from any panics, you can make a simple wrapper function for that.
func try(f func()) {
defer func() {
if err := recover(); err != nil {
fmt.Println("caught panic:", err)
}
}()
f()
}
Then
func main() {
try(f1)
try(f2)
try(f3)
}
I had this code:
defer common.LogWarning(
"b09ee123-f18b-46a8-b80d-f8361771178d:",
resp.Body.Close(), // gets called immediately, is *not* deferred..
)
and common.LogWarning is simply like this:
func LogWarning(uuid string, err error) {
if err != nil {
log.Warning(uuid, err)
}
}
the problem is that resp.Body.Close() gets called immediately - that call is not deferred, so how does this work? Why is not the whole code block get deferred?
From the documentation:
The behavior of defer statements is straightforward and predictable. There are three simple rules:
A deferred function's arguments are evaluated when the defer statement is evaluated.
The defer statement defers the function call. The arguments to the function are evaluated immediately.
Use an anonymous function to accomplish your goal:
defer func() {
common.LogWarning("b09ee123-f18b-46a8-b80d-f8361771178d:",
resp.Body.Close())
}()
The call represented by the trailing () is deferred.
defer defers the execution of a function until the current function returns. The arguments to the function are evaluated immediately.
https://tour.golang.org/flowcontrol/12
If you need to defer a code block where all evaluations are to be deferred, make it a function:
defer func() {
// Stuff to defer here
}()
``
I’m trying to catch crashes/panics from go routines that are created in my program, in order to send them to my crash-error-reporting server (such as Sentry/Raygun)
For example,
func main() {
go func() {
// Get this panic
panic("Go routine panic")
}()
}
The answer states a goroutine cannot recover from a panic in another goroutine.
What would be the idiomatic way to go about it?
You have to "inject" some code into the function that is launched as a new goroutine: you have to call a deferred function in which you call recover(). This is the only way to recover from a panicing state. See related: Why does `defer recover()` not catch panics?
For example:
go func() {
defer func() {
if r := recover(); r != nil {
fmt.Println("Caught:", r)
}
}()
panic("catch me")
}()
This will output (try it on the Go Playground):
Caught: catch me
It is unfeasible to do this in every goroutine you launch, but of course you can move the recovering-logging functionality to a named function, and just call that (but deferred of course):
func main() {
go func() {
defer logger()
panic("catch me")
}()
time.Sleep(time.Second)
}
func logger() {
if r := recover(); r != nil {
fmt.Println("Caught:", r)
}
}
This will output the same (try it on the Go Playground).
Yet another, more convenient and even more compact solution is to create a utility function, a "wrapper" which receives the function, and takes care of the recovering.
This is how it could look like:
func wrap(f func()) {
defer func() {
if r := recover(); r != nil {
fmt.Println("Caught:", r)
}
}()
f()
}
And now using it is even simpler:
go wrap(func() {
panic("catch me")
})
go wrap(func() {
panic("catch me too")
})
It will output (try it on the Go Playground):
Caught: catch me
Caught: catch me too
Final note:
Note that launching an actual goroutine happens outside of wrap(). This gives the caller the option to decide if a new goroutine is required just by prefixing the wrap() call with go. Usually this approach is preferred in Go. This allows you to execute arbitrary functions by passing them to wrap(), and it will "protect" its execution (by recovering from panics, properly logging / reporting it) even if you do not wish to run it concurrently in a new goroutine. On the other hand if you'd move go inside wrap() it wouldn't even work anymore as the recover() call would not happen on the panicking goroutine.
Why does a call to defer func() { recover() }() successfully recover a panicking goroutine, but a call to defer recover() not?
As an minimalistic example, this code doesn't panic
package main
func main() {
defer func() { recover() }()
panic("panic")
}
However, replacing the anonymous function with recover directly panics
package main
func main() {
defer recover()
panic("panic")
}
Quoting from the documentation of the built-in function recover():
If recover is called outside the deferred function it will not stop a panicking sequence.
In your second case recover() itself is the deferred function, and obviously recover() does not call itself. So this will not stop the panicking sequence.
If recover() would call recover() in itself, it would stop the panicking sequence (but why would it do that?).
Another Interesting Example:
The following code also doesn't panic (try it on the Go Playground):
package main
func main() {
var recover = func() { recover() }
defer recover()
panic("panic")
}
What happens here is we create a recover variable of function type which has a value of an anonymous function calling the built-in recover() function. And we specify calling the value of the recover variable to be the deferred function, so calling the builtin recover() from that stops the panicing sequence.
The Handling panic section mentions that
Two built-in functions, panic and recover, assist in reporting and handling run-time panics
The recover function allows a program to manage behavior of a panicking goroutine.
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.
That illustrates that recover is meant to be called in a deferred function, not directly.
When it panic, the "deferred function" cannot be the built-in recover() one, but one specified in a defer statement.
DeferStmt = "defer" Expression .
The expression must be a function or method call; it cannot be parenthesized.
Calls of built-in functions are restricted as for expression statements.
With the exception of specific built-in functions, function and method calls and receive operations can appear in statement context.
An observation is that the real problem here is the design of defer and thus the answer should say that.
Motivating this answer, defer currently needs to take exactly one level of nested stack from a lambda, and the runtime uses a particular side effect of this constraint to make a determination on whether recover() returns nil or not.
Here's an example of this:
func b() {
defer func() { if recover() != nil { fmt.Printf("bad") } }()
}
func a() {
defer func() {
b()
if recover() != nil {
fmt.Printf("good")
}
}()
panic("error")
}
The recover() in b() should return nil.
In my opinion, a better choice would have been to say that defer takes a function BODY, or block scope (rather than a function call,) as its argument. At that point, panic and the recover() return value could be tied to a particular stack frame, and any inner stack frame would have a nil pancing context. Thus, it would look like this:
func b() {
defer { if recover() != nil { fmt.Printf("bad") } }
}
func a() {
defer {
b()
if recover() != nil {
fmt.Printf("good")
}
}
panic("error")
}
At this point, it's obvious that a() is in a panicking state, but b() is not, and any side effects like "being in the first stack frame of a deferred lambda" aren't necessary to correctly implement the runtime.
So, going against the grain here: The reason this doesn't work as might be expected, is a mistake in the design of the defer keyword in the go language, that was worked around using non-obvious implementation detail side effects and then codified as such.
Considering I'm using the original "errors" go package.
And, the difference between panic(11) and panic("11")?
panic is defined as func panic(v interface{}), calling panic(anything) will print the the string representation of anything then the stacktrace of the calling function.
Only difference is, if you use recover, you will be able to access whatever you passed to panic, for example:
func main() {
defer func() {
if err := recover(); err != nil {
if n, ok := err.(int); ok && n == 11 {
fmt.Println("got 11!")
}
}
}()
panic(11)
}
panic("error_msg") and panic("11") panic a string while panic(error.New("error_msg") panics an error and panic(11) panics an integer.
If you do not handle these panics with recover during defer than it won't matter which you use, all will print the "error_msg" or "11".