Floor Function

[franzw82]

I have a question concerning the following code:

real*8 :: floorval
real*8 :: max8int

max8int = 2**63-1
floorval = floor (max8int, 8)

The varibale "floorval" should have the value "2**63-1" but it has the value "-(2**63-1)". I am using compiler version 11.1.051. Am I missing something? Thank you very much in advance!

[Steven_L_Intel1]

You want:

max8int - 2_8**63_8 - 1

The expression 2**63-1 is "default integer" and overflows. The 2**63 ends up as 0 (since all the 1 bits are shifted out) and thus max8int is -1.

I would suggest as an alternative:

max8int = HUGE(0_8)

[franzw82]

Quoting - Steve Lionel (Intel)

You want:

max8int - 2_8**63_8 - 1

The expression 2**63-1 is "default integer" and overflows. The 2**63 ends up as 0 (since all the 1 bits are shifted out) and thus max8int is -1.

I would suggest as an alternative:

max8int = HUGE(0_8)

Thank you for clearing that up!

I have another question concerning the floor function. I have large real number (e.g. around 1.d+20) and would like to compute the floor-values of these numbers. For example:

real*8 :: val = 1.d+20

As far as I understand, the value "floor(val)" does not fit into an integer*8. Is there another way to compute the floor of such a large real number?

Thank you. Franz

[Steven_L_Intel1]

FLOOR returns a REAL value with the same kind of its argument. However, do you understand that when you get to values as large as 1.0D20 that you've lost significance in the low "integer" bits? A REAL*8 can't represent integers larger than 2**52 or thereabouts. So larger than that, there's no point in doing FLOOR as you'll get the same value back.

What exactly are you trying to accomplish here? What is the algorithm?