Loss of precision in java

In the case of int to byte, there's no real concern about a loss of precision, because both types have the same degree of granularity. You'll get an error if you try to convert a literal with a value outside the range of byte to byte. (The error message given in that case is slightly misleading.)

In the case of double to float, you can have a constant value which is in the right range, but still lose precision. In your specific case of 10.0, the value can be represented exactly in both float and double, but that's not the case in general.

As an example of that, consider this:

float f = (float) 10.1; // Or float f = 10.1f;
double d = 10.1;
System.out.println(f == d); // Prints false

That's because precision is being lost in the conversion from double tofloat - neither type can represent 10.1 exactly, but double gets close to it than float does. The == operator will mean f is converted back to a double, with a different value to d.