How to create two variables refering the same nested table? - oracle

Question

I expected that in the following code variables x and y refer the same nested table object but expression y = x seems to create empty table in y. So output is surprisingly 'No: a c' ( however I expected 'Yes: a b c'). What should I do in PL/SQL to have y referring the same nested table object as x.
declare
type str_t is table of varchar(100);
x str_t := str_t ();
y str_t ;
begin
x.extend; x(x.last) := 'a';
y := x;
y.extend; y(y.last) := 'b';
x.extend; x(x.last) := 'c';
dbms_output.put_line(case when x=y then 'Yes: ' else 'No: ' end);
for i in x.first .. x.last
loop
dbms_output.put_line(x(i));
end loop;
Сan it (passing nested table by reference - not by value) be written in PL/SQL or it's fundamentally impossible? Does any alternatives exist in the language to aviod the problem?

Answer 1

PL/SQL doesn't have references (but in one place we'll see later) so essentially what you're asking is impossible in PL/SQL. I don't see how that can be a problem, though.
In your example x and y are two different variables that are having the same type (a nested table collection), x := y is a variable assigment (deep copy), x = y is a comparison. See also e.g. Assigning Values to Variables.
I guess you've programming background with some other languages and you're trying to apply some other language paradigm here. You also might have an XY-problem.
One alternative is to use two different index variables to access the collection. That might or might not suit for your case:
declare
type str_list_t is table of varchar2(32767);
v_foos constant str_list_t := str_list_t('a', 'b', 'c');
-- have some business logic to get the right indices
i pls_integer := 1;
j pls_integer := 3;
begin
dbms_output.put_line(v_foos(i));
dbms_output.put_line(v_foos(j));
end;
/
But please note that changes to collection will invalidate the index variables (that's why I declared a constant collection).
When working with nested tables check also SQL multiset conditions that provide a powerful manipulations.
The only place with reference semantics is subprogram parameters where different parameter modes (IN, OUT or IN OUT) have different semantics (pass-by-value or pass-by-reference). But even there the decision is made by PL/SQL compiler/runtime - programmer has no much control.

Answer 2

So output is surprisingly 'No: a c' ( however I expected 'Yes: a b c').
Your expectation is wrong. Since the two arrays are having different values.
Let's see step by step:
x.extend;
x(x.last) := 'a';
So, x has value a.
y := x;
x is copied to y, so y has value a.
y.extend;
y(y.last) := 'b';
Now, y has a and b.
x.extend;
x(x.last) := 'c';
Now, x has a and c.
Finally,
X --> a,c
Y --> a,b
Let's see what your code shows the output:
SQL> set serveroutput on
SQL> DECLARE
2 type str_t
3 IS
4 TABLE OF VARCHAR(100);
5 x str_t := str_t ();
6 y str_t ;
7 BEGIN
8 x.extend;
9 x(x.last) := 'a';
10 y := x;
11 y.extend;
12 y(y.last) := 'b';
13 x.extend;
14 x(x.LAST) := 'c';
15 FOR i IN x.first .. x.last
16 LOOP
17 dbms_output.put_line('X'||'-'||x(i));
18 END LOOP;
19 FOR i IN y.first .. y.last
20 LOOP
21 dbms_output.put_line('Y'||'-'||y(i));
22 END LOOP;
23 END;
24 /
X-a
X-c
Y-a
Y-b
PL/SQL procedure successfully completed.
SQL>
So, X and Y are not equal.
To make them equal, set Y = X:
SQL> set serveroutput on
SQL> DECLARE
2 type str_t
3 IS
4 TABLE OF VARCHAR(100);
5 x str_t := str_t ();
6 y str_t ;
7 BEGIN
8 x.extend;
9 x(x.LAST) := 'a';
10 x.extend;
11 x(x.last) := 'b';
12 x.extend;
13 x(x.LAST) := 'c';
14 y := x;
15 dbms_output.put_line(case when x=y then 'Yes: ' else 'No: ' end);
16 FOR i IN x.first .. x.last
17 LOOP
18 dbms_output.put_line('X'||'-'||x(i));
19 END LOOP;
20 END;
21 /
Yes:
X-a
X-b
X-c
PL/SQL procedure successfully completed.
SQL>