Function of certification and OCA rest

Hello, I am a student at OCA certification and need help to understand the rest function works.

I read that the remainder (m, n) (http://www.techonthenet.com/oracle/functions/remainder.php) is calculated as: m - (n * X) where X is the closest whole number of m / n.

I ask: why rest (12.5,5) = 2.5?

In my view, the result should be > > > rest (12.5,5) = 12.5-(5 * [l'entier le plus proche de 12.5/5])=12.5-(5*3)=-2.5.)

Thanks for your replies. Kind regards.

antoniob wrote:
It's the right answer... maybe, there seems to be a bug.

You shouldn't jump to conclusions that fast. First let us check the behavior documented. No speculation that can be found on the net (or even on this forum).

Oracle wrote: http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/functions152.htm#sthref1470
...

  • If n1 = 0 or m2 = infinite, then Oracle returns

  • An error if the arguments are of type NUMBER
    NaN if the arguments are of BINARY_FLOAT and BINARY_DOUBLE types.
  • If n1! = 0, then the rest is n2 - (n1 * N) where N is the integer nearest n2/n1. If n2/n1 is equal to x.5, N is the nearest integer pair .

  • If n2 is a floating-point number, and if the remainder is 0, then the sign is the sign of the n2. Remains of 0 are not signed for NUMERIC values.
  • Where you think is out of reminder in conflict with this description?

    Published by: Sven w. on April 7, 2011 16:43

    Tags: Database

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