If you have something like this

   int denom;
   double test;

   test = 1 / denom;

then test will be assigned a value of zero (where denom is not 0). That's
because the division happens in integers, it yields a result of 0 and
integer 0 is converted to double 0.

If on the other hand you have an expression that contains integers and
doubles then the integers are promoted to doubles before the division and
assignment happen.

Is that what you see?

Regards,
Will