Integer division: How do you produce a double?
casting, integer-division, java
Solution
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.
Problem
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.