Hi Chris,

In the expression "(double)(2482884 * 1024)" the cast to double is done
against the result of the expression "(2482884 * 1024)". The "(2482884 *
1024)" expression is evaluated first, and when it is parsed the operands
are two integer values. They are the same type so no implicit conversion
is necessary and the result is a (wrapped) integer value. This value is
then converted to double by the cast.

When you move the conversion inside the parentheses, as you do in your
second two examples, the multiplication expression will have two operands
of type double and the result will be of type double as well, resulting in
no wrap.

Aaron Albertson

c400-l-bounces+albertaa=us.ibm.com@xxxxxxxxxxxx wrote on 06/01/2011
11:17:23 AM:

From: "Chris Hird" <chrish@xxxxxxxxxxxxxxxxx>
To: <c400-l@xxxxxxxxxxxx>
Date: 06/01/2011 03:11 PM
Subject: [C400-L] Conversion to double.
Sent by: c400-l-bounces+albertaa=us.ibm.com@xxxxxxxxxxxx


Not sure why I struggled with this one but.

Printf(“%f\n”,(double)(2482884 * 1024)); shows -1752494080.000000

Printf(“%f\n”,(2482884.0 * 1024.0)); shows 2542473216.000000

Printf(“%f\n”,((double)2482884 * (double)1024)); shows 2542473216.000000

It seems like the conversion to double does not work in the first
Am I doing something wrong? Shouldn’t the cast of the result make the
number correct?



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