|Sep26-12, 05:24 PM||#1|
Another simple fortran q
I want to find a bound for around magnitude 10**270 which is within bounds of double precision but I get
DOUBLE PRECISION :: bound = 10**270
Error: Arithmetic overflow at (1)
where 1 is beneath the first *.
I am sure it is something simple I am overlooking....
|Sep26-12, 05:41 PM||#2|
Since 10 and 270 are both integers, 10**270 is evaluated as an integer expression, and then the result is converted to double precision to assign it to the variable. You got the overflow because 10**270 can't be represetned as an integer value.
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 1038.
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