No need to use a function - you can do it in (Oracle's) SQL.
SELECT LISTAGG(SIGN(BITAND(43, POWER(2,LEVEL-1))),'')
WITHIN GROUP(ORDER BY LEVEL DESC) bin
FROM dual
CONNECT BY POWER(2, LEVEL-1)<=43;
Result:
BIN
101011
I found this deadly snippet here and the fiddle is here. For the number 43, just substitute your column of choice. It's probably possible to do this using recursive CTEs, but that's a bit aboutabove my pay grade :-).
And to reverse the process, you can use this snippet
WITH INPUT AS
(SELECT REVERSE('1000') AS X FROM DUAL)
SELECT SUM(TO_NUMBER(SUBSTR(X,LEVEL,1)*POWER(2,LEVEL-1))) AS OUTPUT
FROM INPUT CONNECT BY LEVEL<=LENGTH(X);
Result:
OUTPUT
8
From here - dbfiddle here. Again, a recursive CTE might do the trick. Again, for '1000', just substitute your column.
Just for kicks, I found another function which will work (with tweaks) on older versions of databases that don't have recursive CTEs. From the excellent Orafaq site here.
SELECT
DECODE(BITAND(VALUE, 128), 128, '1', '0') ||
DECODE(BITAND(VALUE, 64), 64, '1', '0') ||
DECODE(BITAND(VALUE, 32), 32, '1', '0') ||
DECODE(BITAND(VALUE, 16), 16, '1', '0') ||
DECODE(BITAND(VALUE, 8), 8, '1', '0') ||
DECODE(BITAND(VALUE, 4), 4, '1', '0') ||
DECODE(BITAND(VALUE, 2), 2, '1', '0') ||
DECODE(BITAND(VALUE, 1), 1, '1', '0') as bin_number from
(select 8 as value from dual) A;
Result:
MY_BIN
1000
Fiddle herehere. The more observant among you will notice that that code does not remove leading zeros - the code above has to be encase (entombed?) in this snippet
select replace(ltrim(replace(ColumnName,'0',' ')),' ','0')
which can be found here.