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.

Answer 1

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.