Another simple fortran q

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.