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.
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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.