This arithmetic predicate calculates the exponential e to it's argument. If an overflow is detected the exception evaluation_error(float_overflow) is thrown.
see also: e/0 arithmetic in Prolog is/2
| Examples | |
| X is exp(1.1). | succeeds, X is 3.004166 |
| X is exp(-100). | an evaluation_error(float_overflow). exception is thrown |
| X is exp(-10). | succeeds, X is 4.539993e-05 |
| Exceptions | |
| The result exceeds the floating point range | an evaluation_error(float_overflow) exception is thrown |