Because I wanted an UUID I played around with sys_guid() and stumbled upon something that was new for me.

select substr(l,1,8) ||'-'||substr(l,9,4)||'-'
||substr(l, 13, 4)||'-'||substr(l,17, 4)||'-'
||substr(l,21, 12) as final
, l
from
( select sys_guid() as l
from dual
) sub

The statement gave two different results!

So sub.l is not equal to sub.final minus the minus signs.

But when I do this:

select substr(l,1,8) ||'-'||substr(l,9,4)||'-'
||substr(l, 13, 4)||'-'||substr(l,17, 4)||'-'
||substr(l,21, 12) as final
, l
from
( select sys_guid() as l
from other_table_with_1_row
) sub

The results are equal.

My hypothesis was that the fast dual rewrite replaces every occurrence of sub.l with sys_guid(). So, sys_guid is handled six times as a constant, but of course it isn’t.

drop sequence edwin_seq;
create sequence edwin_seq start with 1 increment by 1 nocache;
drop function edwin_f;
create function edwin_f
return number
is
begin
return edwin_seq.nextval;
end;
select l + l as sum
, l
from
( select edwin_f as l
from other_table_with_1_row
) sub

The resultset is that the sub.sum equals 2 and sub.l equals to 1.

But now do the same thing with dual:

drop sequence edwin_seq;
create sequence edwin_seq start with 1 increment by 1 nocache;
drop function edwin_f;
create function edwin_f
return number
is
begin
return edwin_seq.nextval;
end;
select l + l as sum
, l
from
( select edwin_f as l
from dual
) sub

Now sub.sum equals 3 and sub.l also equals to 3.

This is because the internal algorithm handles the function as a constant and the resultset is calculated as

first-column = edwin_f + edwin_f = 1 + 2 = 3
second-column = edwin_f = 3

Every time the function is called the next value of the sequence is returned.

So, be careful with the use of dual in combination with functions that do not return constant values.