// Question 2. What will be displayed on the screen package main import "fmt" func main() { { defer fmt.Println(1) } defer fmt.Println(2) panic("aaaa") defer func(){ if r := recover(); r != nil{ fmt.Println("Panic handled", r) } }() }

This code will compile, but when run, it will cause a panic and not produce the expected result due to the order of defer and panic calls.

Analysis:

• Inside the block { defer fmt.Println(1) }, the print of 1 is deferred.
• Then, the print of 2 is deferred.
• Then, panic("aaaa") is called.
• After the panic, the last defer with a recover function is called.

However, in Go, defer statements are executed in reverse order of their declaration when exiting a function, including when exiting due to a panic.

Order of defer execution during panic:

1. The last defer — the function with recover — is executed.
2. If recover succeeds, the panic is stopped, and the program continues.
3. Then, the remaining defers — fmt.Println(2) and fmt.Println(1) — are executed.

But in this code, the defer with recover is declared after the panic, so it will not be called, as panic interrupts execution immediately.

Actually, the code will not compile because defer statements after a panic are invalid — panic interrupts execution, and subsequent code is not executed.

If the defer with recover had been declared before the panic, the output would be:

Panic handled aaaa
2
1

Summary: In its current form, the code will cause a panic and terminate with an error, outputting nothing, because the defer with recover is declared after the panic and will not execute.