How to delete elements by using recursion? - ada

Question

In this code, I have built a list of three integers (5, 10, 15) and what I need help with is that I need to ask the user which of these elements he/she wants to remove and then only return the element/elements that are left. I need to write a subprogram for this and just by using recursion, I need to remove the elements that the user does not need.
Main program:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
with Linked_List; use Linked_List;
procedure Main_Remove is
I : Integer;
L : List_Type;
begin
Build_Test_List(L); -- builds a list of 3 integers (5 => 10 => 15)
Put(L);
Put("Which elements do you want to remove/delete ");
Get(I);
Remove(L, I);
Put(L);
end Main_Remove;
Package:
package Linked_List is
type List_Type is private;
procedure Put(Item : in List_Type);
procedure Build_Test_List(Item : out List_Type;
Base : in Integer := 5);
private
type E_Type;
type List_Type is access E_Type;
type E_Type is
record
Data : Integer;
Next : List_Type;
end record;
end Linked_List;
Pakage body:
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
with Ada.Unchecked_Deallocation;
package body Linked_List is
procedure Put(Item : in List_Type) is
P : List_Type := Item;
begin
Put("Listan: ");
while P /= null loop
if P /= Item then
Put(" -> ");
end if;
Put(P.Data, Width => 0);
P := P.Next;
end loop;
New_Line;
end Put;
procedure Insert_First(L : in out List_Type;
D : in Integer) is
begin
L := new E_Type'(Data => D, Next => L);
end Insert_First;
procedure Build_Test_List(Item : out List_Type;
Base : in Integer := 5) is
begin
for I in reverse 1..3 loop
Insert_First(Item, Base * I);
end loop;
end Build_Test_List;
end Linked_List;

Answer 1

Something like this will do, with reservations: specifically, there’s a memory leak.
procedure Remove (L : in out List_Type; Item : Integer) is
begin
The recursion has to be stopped when the list is empty.
if L = null then
return;
end if;
The list isn’t empty. What to do next depends on whether the current list element contains the value we’re looking for, or not.
if L.Data = Item then
This item needs to be removed from the list. Do this by altering the original pointer (which came from the list head, or the previous element) to skip over this element, and then process that element.
This is the point at which the memory leak has occurred. Obviously the cell being pointed to by the initial L needs to be freed, but you’re going to have to be careful about the order of operations.
L := L.Next;
Remove (L, Item);
else
The item stays in the list, go on to the next element.
Remove (L.Next, Item);
end if;
end Remove;