I can't to understand, why this program prints 421 instead of 431?
package main
import "fmt"
var x int
func f() int {
x
return x
}
func main() {
o := fmt.Println
defer o(f())
defer func() {
defer o(recover())
o(f())
}()
defer f()
defer recover()
panic(f())
}
Below I added the comment how I guess:
package main
import "fmt"
var x int
func f() int {
x
return x
}
func main() {
o := fmt.Println
defer o(f()) // x=1
defer func() {
defer o(recover()) // x=3 from panic (but if we ll execute this program we ll see 2)
o(f()) // x=4
}()
defer f() // x=2
defer recover() //nil
panic(f()) // x=3
}
CodePudding user response:
defer does not call the function, but it does evaluate its arguments "immediately". Also a call to recover() only stops the panicing state if it gets called from a deferred function (defer recover() does not qualify for this). See Why does `defer recover()` not catch panics?
In the light of this: Let's number the lines:
L1: o := fmt.Println
L2: defer o(f()) // x = 1
L3: defer func() {
L4: defer o(recover()) // recover() returns 2
L5: o(f()) // x = 4, it gets printed
L6: }()
L7: defer f() // after panic: x = 3
L8: defer recover()
L9: panic(f()) // x = 2
The execution of the above code will go like this:
L2: evaulates the params of o(), f() is called, x is incremented to 1 (this will be printed later). o() is not yet called.
L3: Deferred function is not called yet, skip its whole body for now.
L7: f() is not called yet, x remains 1.
L8: recover() is not called.
L9: f() is called, increments x to 2, and returns it, so 2 is passed to panic().
We're in a panicking state, so deferred functions are executed now (in LIFO order).
L8: recover() is called but does not stop the panicing state.
L7: f() is called now, increments x to 3.
L3: This anonymous function is now executed.
L4: recover() returns 2 (the value that was passed to panic()), this will be printed later, but not yet, as call to o() is deferred. Panicing state stops here.
L5: f() is called, increments x to 4, it gets printed right away.
L4: deferred function o() is now executed, printing the above value 2.
L2: deferred function o() is now executed, printing the previously evaluated value 1.
End of program.
