Not in this example. But sometimes it does, when values are close to the limit of what can be represented in 16 bits.
Here is another example:
short x = 12; short y = 3; short value; value = x / y;
x / y divides a
32-bit 12 by a 32-bit 3, even though the variables
y are only 16 bits wide.
The calculation produces a 32-bit result.
Because the 32-bit result does not fit in the 16 bits of
the compiler will not compile the last statement.
This can be completely baffling when it happens to you.
C:/Private>javac ShortStuff.java ShortStuff.java:9: possible loss of precision found : int required: short result = x / y; ^ 1 error C:/Private>
At the professional programming level, details like these are sometimes important.
But mostly for general purpose programs use
long for integers and
double for floating point.
This will keep you out of trouble.
If you can't avoid the problem, use a type cast, as described in chapter 28.
Do you expect that a modern electronic calculator will give you the same
answer as Java for the expression
(31.5 - 12)/4.1