Re: Rounding off double precision

glen herrmannsfeldt <gah@xxxxxxxxxxxxxxxx> writes:

Bamm wrote:
(snip)

However I have since found where I went wrong. I was using

double precision Speed, Azimuth, V
Azimuth = 90.
parameter(Pi = 3.141592653589793238d0)
V = Speed * cos(Azimuth * Pi / 180.)
print *, V

The problem with the unusually large value of -4.37113886E-06 arose
because I didn't declare Pi to be double precision, even though I
assigned a double precision value to it.

This problem might be more likely to occur to ex-C programmers.

C uses #define is places where Fortran uses PARAMETER,
but #define is typeless.

There's a little more to it than that. In C, the literal constant "0.1"
has type double precision, in Fortran, the literal constant "0.1" has
type single precision. So

double x = 0.1;

in C gives x the double precision value

+1.600000000000000088817841970012523233890533447265625 / 16

whereas

DOUBLE PRECISION X
X = 0.1

in Fortran gives X the single precision value

+1.60000002384185791015625 / 16

Actually, now that I look at it, the OP actually had the opposite
problem, analogous to

X = 0.1D0

where X is implicitly typed default real (single precision), but the
literal constant has type double precision.

Chip

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