What's the difference between declaring a long variable with 'L' and without?

Question

What's the difference between the following?

long var1 = 2147483647L;

long var2 = 2147483648;

(or any primitive variable declaration) Does it have any performance issue with or without L? Is it mandatory?

Answer 1

In the first case you are assigning a long literal to a long variable (the L or l suffix indicates long type).

In the second case you are assigning an int literal (that's the default type when no suffix is supplied) to a long variable (which causes an automatic type cast from int to long), which means you are restricted to the range from Integer.MIN_VALUE to Integer.MAX_VALUE (-2147483648 to 2147483647).

That's the reason why

long var2 = 2147483648;

doesn't pass compilation (2147483648 is larger than Integer.MAX_VALUE).

On the other hand

long var2 = 2147483648L;

would pass compilation.

Answer 2

For easy understanding each of the type have range in java.

1. By default every digit you entered in java is either byte or short or integer.

   short s = 32767;
   byte b = 127;
   int i = 2147483647;
   

So if you assign anything except from their range you'll get compilation error.

int i = 2147483648;  //compilation error.

Type range

And when you write long longNumber = 2147483647;

though it falls in long range but internally java treat it as

long l = (int) 2147483647;

you wont get any errors.

But if we assign beyond the range of integer like longNumber = 2147483648; we will get compilation error as

long o = (int) 2147483648; 

here java will try to convert the 2147483648 to int but it is not in int range so widening error is thrown. To indicate java that the number what we have written is beyond the integer range just append l or L to the end of the number. so java will wide his range till long and convert it as

long o = (long) 2147483648; 

2. By default every floating point or digit with floating points (.) are size of double. So when you write some digits with (.) java treat as a double and it must be in double range. As we know the float range is smaller then double. so when you write

   float f = 3.14;

though it falls in double range but internally java treat this assignment as

float f = (double) 3.14;

here you are assigning the double to float narrowing which is not correct. so either you have to convert the expression like that

float f = (float)3.14;

or

float f = 3.14f; // tell jvm  to assign this in float range by appending **f** or **F**

Answer 3

If we don't mention the L with the value then value is considered to be int value.

It type casts the int to long automatically.