Reg: concept of COUNT and NULLs -

Question

Hi Experts,
I have a simple conceptual doubt regarding COUNT and NULLs.
Even I have experimented and can justify the behaviour, just need to confirm once with guru's and experts over here.
Workout:
CREATE TABLE test_x1
AS
SELECT 1 col1, 'a' col2, 'a1' col3 FROM dual UNION ALL
SELECT 2 col1, 'a' col2, 'a1' col3 FROM dual UNION ALL
SELECT 3 col1, 'a' col2, 'a1' col3 FROM dual UNION ALL
SELECT 4 col1, 'a' col2, 'a1' col3 FROM dual UNION ALL
SELECT 5 col1, 'a' col2, 'a1' col3 FROM dual UNION ALL
SELECT 6 col1, 'a' col2, NULL col3 FROM dual UNION ALL
SELECT NULL col1, NULL col2, NULL col3 FROM dual UNION ALL
SELECT 8 col1, 'a' col2, NULL col3 FROM dual;
SELECT
  Count(*),
  Count(col1),
  Count(col2),
  Count(col3)
FROM test_x1;
COUNT(*) COUNT(COL1) COUNT(COL2) COUNT(COL3)
       8           7           7           5
My justification:
COUNT(*) - It counts for physical *rowid* ,and doesn't bother to check if data in the columns is it NULL or anything else.
Whereas, COUNT(<column>) - just scans/counts the known values and NOT NULLs.
Is the above correct? Please advise me on this.
-- Ranit

Answer 1

it's documented: http://docs.oracle.com/cd/E11882_01/server.112/e41084/functions039.htm#SQLRF00624
If you specify expr, then COUNT returns the number of rows where expr is not null. You can count either all rows, or only distinct values of expr.
If you specify the asterisk (*), then this function returns all rows, including duplicates and nulls. COUNT never returns null.
So the answer is: yes.