Using the letter L in long variable declaration

Question

long l2 = 32;

When I use the above statement, I don't get an error (I did not used l at the end), but when I use the below statement, I get this error:

The literal 3244444444 of type int is out of range

long l2 = 3244444444;

If I use long l2 = 3244444444l;, then there's no error.

What is the reason for this? Using l is not mandatory for long variables.

Answer 1

3244444444 is interpreted as a literal integer but can't fit in a 32-bit int variable. It needs to be a literal long value, so it needs an l or L at the end:

long l2 = 3244444444l; // or 3244444444L

More info:

• Primitive Data Types, specifically Default Values and Literals sections.

Answer 2

Note that while int literals will be auto-widened to long when assigning to a long variable, you will need to use an explicit long literal when expressing a value that is

1. greater than Integer.MAX_VALUE (2147483647)

   (or)

2. less than Integer.MIN_VALUE (-2147483648):

   long x1 = 12; //OK
   long x2 = 2147483648; // not OK! That's not a valid int literal
   long x3 = 2147483648L; // OK