// 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 recovered", r) } } }

This code has several important points:

1. defer fmt.Println(1) is called inside a block, but this does not affect the order of deferred calls — they are added to a stack.
2. Then defer fmt.Println(2) is called.
3. After that, panic("aaaa") is called.
4. After panic, there is a defer with a function that tries to perform recover().

However, the defer func(){...} is declared after the panic, so this defer will never be registered because the function execution will be interrupted at panic.

Execution order:

• When panic is called, all deferred calls are executed in reverse order of their registration.
• First, defer fmt.Println(2) executes — outputs 2.
• Then defer fmt.Println(1) executes — outputs 1.
• defer func(){...} is not registered because it is declared after panic.
• The panic is unhandled, and the program will terminate with an error.

The output on the screen will be:

2
1
panic: aaaa

If defer func(){...} had been declared before panic, it would have caught the panic and printed "Panic recovered aaaa".