Why this program prints 421 in result?

Question

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
}

[How to debug small programs](https://ericlippert.com/2014/03/05/how-to-debug-small-programs/) — Filburt, Jan 11 '23 at 10:42

icza · Accepted Answer · 2023-01-11T12:21:36.167

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.

Why this program prints 421 in result?

1 Answers1