Question

I came across the following SQL statement:

    SELECT A.NAME
    FROM   THE_TABLE A
    WHERE  A.NAME LIKE '%JOHN%DOE%'
    AND    ((A.NUM_FIELD/1) - (A.NUM_FIELD/2)*2 <> 0)

That last condition, "((A.NUM_FIELD/1) - (A.NUM_FIELD/2)*2 <> 0)" is what baffles me. Depening on the implementation of order of operations, it should always result to 0 or A.NUM_FIELD / 2.

How does SQL still return records from this view? If it always results to half the original value, why have it? (This is a delivered SQL package)

Answer 1

What SQL implementation is this?

As noted,

(x/1) - (x/2)*2

is equivalent to

X - (2*(x/2))

which, if integer division is being performed yields 0 or 1 depending on whether the value is even or odd:

x x/2 2*(x/2) x-(2*(x/2))
- --- ------- -----------
0   0       0           0
1   0       0           1
2   1       2           0
3   1       2           1
4   2       4           0
...

if so, it seems like an odd way way of checking for odd/even values, especially since most SQL implementations that I’m aware of support a modulo operator or function, either x % y or mod(x,y).

The seemingly extraneous division by 1 makes me think the column in question might be floating point. Perhaps the person who coded the query is trying to check for jitter or fuzzyness in the low order bits of the floating point column?

if you modified the query to return all the intermediate values of the computation:

SELECT A.NAME                    as Name      ,
       A.NUM_FIELD               as X         ,
       A.NUM_FIELD / 1           as X_over_1  ,
       A.NUM_FIELD / 2           as X_over_2  ,
       ( X.NUM_FIELD / 2 )
       * 2                       as 2x_over_2 ,
         ( A.NUM_FIELD / 1 )
       - ( A.NUM_FIELD / 2 ) * 2 as Delta ,
       case when ( ( A.NUM_FIELD / 1 ) - ( A.NUM_FIELD / 2 ) * 2 ) <> 0
       then 'return'
       else 'discard'
       end as Row_Status
FROM   THE_TABLE A
WHERE  A.NAME LIKE '%JOHN%DOE%'

and then executed it, what results do you get?