- #1

- 68

- 0

DOUBLE PRECISION :: bound = 10**270

Error: Arithmetic overflow at (1)

.\q2.f03:70.4:

where 1 is beneath the first *.

I am sure it is something simple I am overlooking....

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- Fortran
- Thread starter autobot.d
- Start date

- #1

- 68

- 0

DOUBLE PRECISION :: bound = 10**270

Error: Arithmetic overflow at (1)

.\q2.f03:70.4:

where 1 is beneath the first *.

I am sure it is something simple I am overlooking....

- #2

AlephZero

Science Advisor

Homework Helper

- 7,002

- 294

You could write 10.0d0 ** 270, or better, just 10.0d270.

Note you must use a "d" exponent (not "e") in a constant to make it double precision. 10.0 ** 270 will still give an overflow, because 10.0 is a single precision constant and the maximum value that can be represented in single precision is about 10

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