# Is There a Better Double-Precision Assignment in Fortran 90

fortranfortran90

In Fortran 90 (using gfortran on Mac OS X) if I assign a value to a double-precision variable without explicitly tacking on a kind, the precision doesn't "take." What I mean is, if I run the following program:

``````program sample_dp

implicit none

integer, parameter :: sp = kind(1.0)
integer, parameter :: dp = kind(1.0d0)

real(sp) :: a = 0.
real(dp) :: b = 0., c = 0., d = 0.0_dp, e = 0_dp

! assign values
a = 0.12345678901234567890
b = 0.12345678901234567890
c = DBLE(0.12345678901234567890)
d = 0.12345678901234567890_dp

write(*,101) a, b, c, d
101 format(1x, 'Single precision: ',  T27, F17.15, / &
1x, 'Double precisison: ', T27, F17.15, / &
1x, 'Double precision (DBLE): ', T27, F17.15, / &
1x, 'Double precision (_dp): ',  T27, F17.15)

end program
``````

I get the result:

``````Single precision:        0.123456791043282
Double precision:        0.123456791043282
Double precision (DBLE): 0.123456791043282
Double precision (_dp):  0.123456789012346
``````

The single precision result starts rounding off at the 8th decimal as expected, but only the double precision variable I assigned explicitly with _dp keeps all 16 digits of precision. This seems odd, as I would expect (I'm relatively new to Fortran) that a double precision variable would automatically be double-precision. Is there a better way to assign double precision variables, or do I have to explicitly type them as above?

#### Best Solution

A real which isn't marked as double precision will be assumed to be single precision. Just because sometime later you assign it to a double precision variable, or convert it to double precision, that doesn't mean that the value will 'magically' be double precision. It doesn't look ahead to see how the value will be used.