Java - int/long, float/double

double, floating-point, java, long-integer, primitive-types

Solution

12 as a constant in java is an int.

The reason long l = 12 compiles is that it is automatically widened to a long.

EDIT: Regarding your comment, there is nothing wrong with automatic widening, use whatever makes your code clearer, but just be aware of what is going on when you do math on primitives. For example:

int i = 1213;
long l = 112321321L * i;

The long will have a very different value if you don't explicitly put that first number as a long because Java will treat it as integer math, and cause an overflow.

Problem

I understand that "2.5" is automatically a double, and to make it a float, I need to do "2.5F" (or should the F be lowercase?), and that I should use a float, say, if I had a constant that only ever needed 2 decimal spaces (like "0.08F" for Ontario PST tax), but I'm not sure whether "12" is an int or a long, but I know "12L" is a long, but "long l = 12" and "long l = 12L" seem to compile to the same thing, and I use long's if I want maximum non-floating point precision, and int if I know I won't need beyond int's limits. Please correct me if there's something there that isn't right, and answer the quesions I have.

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