I have what I imagine is a common use case for localprec(): I have a calculation which is likely to need additional precision so it can return a result at the current precision. My inclination was to write
foo()=localprec(default(realprecision)+9); ...
but I realized that default(realprecision) is really the wrong thing, as it could be higher or lower than the current precision, as indeed ??localprec states. Is there a good way to do this?
Charles Greathouse
Analyst/Programmer
Case Western Reserve University