Integer division How do you produce a double

Question

For this code block   int num   5  int denom   7  double d   num   denom    the value of d is 0 0  It can be forced to work by casting   double d     double  num    denom    But is there another way to get the correct double result  I don t like casting primitives  who knows what may happen

User · Answer

just use this   int fxd 1  double percent   double  fxd 40  100

User · Answer

You might consider wrapping the operations  For example   class Utils       public static double divide int num  int denom            return   double  num    denom            This allows you to look up  just once  whether the cast does exactly what you want  This method could also be subject to tests  to ensure that it continues to do what you want  It also doesn t matter what trick you use to cause the division  you could use any of the answers here   as long as it results in the correct result  Anywhere you need to divide two integers  you can now just call Utils  divide and trust that it does the right thing

User · Answer

Just add  quot D quot   int i   6  double d   i   2D     This will divide bei double  System out println d      This will print a double    3D

User · Answer

use something like   double step   1d   5     1d is a cast to double

User · Answer

I don t like casting primitives  who knows what may happen    Why do you have an irrational fear of casting primitives  Nothing bad will happen when you cast an int to a double  If you re just not sure of how it works  look it up in the Java Language Specification  Casting an int to double is a widening primitive conversion   You can get rid of the extra pair of parentheses by casting the denominator instead of the numerator   double d   num    double  denom

User · Answer

Type Casting Is The Only Way May be you will not do it explicitly but it will happen  Now  there are several ways we can try to get precise double value  where num and denom are int type  and of-course with casting -  with explicit casting    double d    double  num   denom  double d     double  num    denom  double d   num    double  denom  double d    double  num    double  denom   but not double d    double   num   denom    with implicit casting    double d   num   1 0   denom  double d   num   1d   denom  double d     num   0 0     denom  double d   num     d    denom   but not double d   num   denom   1 0  and not double d   0 0     num   denom     Now if you are asking- Which one is better  explicit  or implicit  Well  lets not follow a straight answer here  Simply remember- We programmers don t like surprises or magics in a source  And we really hate Easter Eggs  Also  an extra operation will definitely not make your code more efficient  Right

User · Answer

If you change the type of one the variables you have to remember to sneak in a double again if your formula changes  because if this variable stops being part of the calculation the result is messed up  I make a habit of casting within the calculation  and add a comment next to it   double d   5    double  20    cast to double  to do floating point calculations   Note that casting the result won t do it  double d    double  5   20     produces 0 0

User · Answer

What s wrong with casting primitives   If you don t want to cast for some reason  you could do  double d   num   1 0   denom

User · Answer

Cast one of the integers both of the integer to float to force the operation to be done with floating point Math  Otherwise integer Math is always preferred  So   1  double d    double 5   20  2  double v    double 5    double  20  3  double v   5    double  20    Note that casting the result won t do it  Because first division is done as per precedence rule   double d    double  5   20     produces 0 0   I do not think there is any problem with casting as such you are thinking about

User · Answer

Best way to do this is  int i   3  Double d   i   1 0   d is 3 0 now

User · Answer

double num   5    That avoids a cast   But you ll find that the cast conversions are well-defined   You don t have to guess  just check the JLS   int to double is a widening conversion   From   5 1 2      Widening primitive conversions do not   lose information about the overall   magnitude of a numeric value                  Conversion of an int or a long value   to float  or of a long value to   double  may result in loss of   precision-that is  the result may lose   some of the least significant bits of   the value  In this case  the resulting   floating-point value will be a   correctly rounded version of the   integer value  using IEEE 754   round-to-nearest mode    4 2 4     5 can be expressed exactly as a double

[java] Integer division: How do you produce a double?

Examples related to java

Examples related to casting

Examples related to integer-division