Question

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All literals defined in code are by default integers unless the literal is a float or a double. Floats are denoted with an L or l and the doubles with a D or d.

What is the consequence of this and how does it affect casting?

Answer

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The first issue to consider is the maximum and minimum value of an integer. No integer literal can out of the range of an integer. The range is from -2,147,483,648 to 2,147,483,647 (inclusive). Code that contains an integer out of this range will not compile.

The second issue relates to explicit and implicit casting. I will demonstrate with an example:

[sourcecode language=”java” toolbar=”false”]
short s1 = 32767;
short s2 = (short) 32768;
short s3 = (short) 2147483648;
short s4 = (short) 2147483648f;
short s5 = (short) 2147483648d;
[/sourcecode]

A short has a range from -32,768 to 32,767 (inclusive) therefore in:

line 1 the literal is implicitly cast to a short because it is within the range of a short.
line 2 the literal is outside the range of a short (but within the range of a integer) and therefore must be explicitly cast down to a short. There will be a loss of precision.
line 3 the literal is outside the range of an integer and therefore this code will not compile.
line 4 the literal is within the range of a float and is implicitly cast to an integer then explicitly cast to a float. Notice the f and the end of the number.
line 5 the literal is a double and with the range of a double. Notice the d and the end of the number.

As you can see it can be tricky to work with literals in the code.