Explain the use of Go's variable shadowing and redeclaration in nested scopes?

In Go, variable shadowing occurs when a variable with the same name as an outer-scoped variable is declared in an inner scope, effectively "hiding" the outer-scoped variable within that inner scope. This can be useful in cases where you need to temporarily modify the value of a variable without affecting the outer scope.

For example, consider the following code:

package main

import "fmt"

func main() {
    x := 10
    fmt.Println(x) // prints 10

    if x := 5; x < 20 {
        fmt.Println(x) // prints 5
    }

    fmt.Println(x) // prints 10
}

In this example, we declare a variable **x** with the value 10 in the outer scope, and then declare another variable **x** with the value 5 in the inner scope of the **if** statement. This inner-scoped variable **x** "shadows" the outer-scoped variable **x** within the **if** statement, so when we print the value of **x** within the **if** statement, we get 5 instead of 10. However, once we exit the **if** statement and return to the outer scope, the outer-scoped variable **x** is once again in scope and its value of 10 is printed.

Redeclaration, on the other hand, occurs when a variable with the same name is declared multiple times in the same scope. This is not allowed in Go, and will result in a compilation error. However, if the redeclaration occurs in nested scopes, it is allowed as long as each declaration is in a separate scope. For example:

package main

import "fmt"

func main() {
    x := 10
    fmt.Println(x) // prints 10

    if y := 5; x < 20 {
        fmt.Println(x, y) // prints 10 5

        x := 20
        fmt.Println(x, y) // prints 20 5
    }

    fmt.Println(x) // prints 10
}

In this example, we declare the variable **x** with the value 10 in the outer scope, and then declare the variable **y** with the value 5 in the inner scope of the **if** statement. We then redeclare the variable **x** with the value 20 in the inner scope, which is allowed because it is in a separate scope from the outer-scoped **x** variable. When we print the values of **x** and **y** within the **if** statement, we get 10 and 5 respectively, because the inner-scoped **x** has not yet been initialized. Once we initialize the inner-scoped **x** to 20, we print the values of **x** and **y** again and get 20 and 5 respectively. Finally, we print the value of the outer-scoped **x** and get 10.