double dAux; ... dAux=Math.round((dAux*1000)); dAux/=1000; // diferent result dAux=(Math.round((dAux*1000)))/1000; ...
posted 14 years ago
Originally posted by David Cuesta: double dAux; ... dAux=Math.round((dAux*1000)); dAux/=1000; // diferent result dAux=(Math.round((dAux*1000)))/1000; ...
I believe what you're seeing is caused by the fact that Math.round returns an int, rather than a double. This is causing integer division (using truncation) to be done instead of floating point division. I used the following code:
and I got the results: 12.6 12.0 However, if I modify the code slightly to look like this:
I then get this as my output: 12.6 12.6 In the second case, Because Math.round(double) returns an int, both operands are inetegers and integer division is being performed. In the first case, since dAux is a double, floating point division takes place automatically (because one of the two operands is a double). You need to specifiy explicitly (through the use of a cast) what type of division you want to occur. By casting the result as a double in the second case, I'm telling the JVM that I want it to perform this division but, even though both operands are integers, I want to use floating point division and avoid the truncation you were originally seeing. I hope this all makes sense. Corey [This message has been edited by Corey McGlone (edited December 27, 2001).]