Duplicating a list to a list with pointers

I have simplified the problem incredibly to a small runnable example of my issue, shown below:

package main

import "fmt"

type A struct {
	Name string
}

func main() {
	main_list := []A{A{"1"}, A{"b"}, A{"3"}}
	second_list := make([]*A, 0)
	fmt.Println("FIRST LIST:")
	for _, x := range main_list {
		fmt.Println(x.Name)
		second_list = append(second_list, &x)
	}
	fmt.Println("SECOND LIST:")
	for _, x := range second_list {
		fmt.Println((*x).Name)
	}
}

Which provides:

FIRST LIST:
1
b
3
SECOND LIST:
3
3
3

The simple task is creating main_list with some dummy structs. The real issue is creating references (pointers) from the values in main_list into second_list. I absolutely do not want copies of the structs, only pointers/references to the main_list structs. The second for loop iterates through the newly populated second_list and shows only the last value from main_list (3 in this example) three times.

I expect the issue arrises with the way I am using &x in the first loop. With the way I get all three values of second_list to all be the same struct instance; I am going to assume that I actually made a pointer to the for-loop's iterator (as if it were a reference?). So all the pointers in second_list will always be the last item referenced in the first loop.

The question: How could I make a pointer from what x was pointing to at that moment in the for loop?

You're adding the address of the same x in every call to append.

You could initialize a new x and copy the value:

for _, x := range main_list {
	x := x
	second_list = append(second_list, &x)
}

Or create a new x directly indexing the slice:

for i := range main_list {
	x := main_list[i]
	second_list = append(second_list, &x)
}

Or if you want the address of the original value in the slice, you can use:

for i := range main_list {
	second_list = append(second_list, &main_list[i])
}