I have the following code in FORTRAN 77:
REAL*8 :: dm dm=1.-1.E-12 write(6,*) 'dm: ', dm I get: dm: 1
Is this OK? I would like to get dm=0.999999999999
- I don't know enough about FORTRAN to answer authoritatively, but assuming that REAL*8 represents an 8 byte float (double precision) it should have at least 15 digits of precision. Could it have something to do with the output formatting?axblount– axblount2013-06-07 20:15:43 +00:00Commented Jun 7, 2013 at 20:15
- 3The trouble is your constants are single precision, try 1.D0-1.D-12. Note that notation (along with REAL*8) is obsolete in modern fortran but still works. Not clear if/why you want f77..agentp– agentp2013-06-07 20:21:07 +00:00Commented Jun 7, 2013 at 20:21
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1 Answer
As stated in a comment, you need to specify the precision of the constants. Also, real*8 is obsolete. (Was it always an extension?) Here is a modern way to write this, using the ISO Fortran Environment to obtain a 64-bit real type and using that type both in the declaration and in the constants.
use ISO_FORTRAN_ENV real (real64) :: dm dm = 1.0_real64 - 1.0E-12_real64 For more info, see What does `real*8` mean?
