func updateV1(px *int) { x2 := 20 px = &x2 } func updateV2(px *int) { *px = 20 } func main() { x := 10 updateV1(&x) fmt.Println(x) updateV2(&x) fmt.Println(x) }

This code contains two functions that accept a pointer to an int but work differently:

func updateV1(px *int) {
    x2 := 20
    px = &x2
}

func updateV2(px *int) {
    *px = 20
}

func main() {
    x := 10
    updateV1(&x)
    fmt.Println(x) // Will print 10
    updateV2(&x)
    fmt.Println(x) // Will print 20
}

Explanation:

• In updateV1, the parameter px is a local copy of the pointer. Assigning px = &x2 only changes the local copy and does not affect the variable x in main.
• In updateV2, dereferencing the pointer *px = 20 changes the value at the address px points to. This modifies the value of x in main.

Result: updateV1 does not change the original variable, but updateV2 does.

This demonstrates the difference between modifying a local copy of a pointer and changing the value at the address it points to.