Rene Brun wrote: > > This is the expected result. > Replace > a = (Double_t)((b-c)/b) > by > a = (Double_t)(b-c)/b Maybe you should explain... (Double_t)((b-c)/b) means: convert the result of the calculation (b-c)/b to double. But since b and c are both integers, the result will be integer too. And since b-c is less than b the outcome is zero. An integer zero of course will be transformed to a double zero. But when you covert b-c to double BEFORE the division, the result will be a double. Now you have a double (b-c) mixed with an integer (b), in such a case the result will always be of the more precise type. Ingo Strauch ---------------------------------------------------------------------------- Ingo Strauch Budapester Str. 30, 20359 Hamburg, Tel.: +49 40 7421 4722 Office: DESY, Group H1/AAC1, Room 1c354, Notkestr. 85, 22607 Hamburg strauch@mail.desy.de Tel. +49 40 8998 2312 http://www-h1.desy.de/~strauch/ FAX +49 40 8998 4385 I. Phys. Institut RWTH Aachen, Room 28B210, Sommerfeldstr. 14, 52056 Aachen strauch@toots.physik.rwth-aachen.de Tel. +49 241 80 7183 http://www-users.rwth-aachen.de/Ingo.Strauch/ FAX. +49 241 8888 661 ----------------------------------------------------------------------------
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