Java - Is There Any Reason to Use This Format: (long) 0; Instead of This One: 0L;?

Question

I couldn't find any information about this when searching StackOverflow or Google, and I've got some coworkers who disagree with my preference for how to initialize a simple Long variable.

Is there any reason to use one of these formats over the other?

Long foo = (long) 0; // First way of doing it
Long bar = 0L; // Second way of doing it

I'm most interested if anyone knows if there is an efficiency difference here.

The reason I prefer the second way is because you can specify values less than Integer.MIN_VALUE and greater than Integer.MAX_VALUE, whereas Eclipse would complain with something along the lines of "The literal 10000000000 of type int is out of range" if you used the first way here.

Answer 1

There is no difference (except you mentioned). Compiler is smart enough. If I compile following class:

public class Test {
    public static void main(String[] args) {
        Long foo = (long) 0;
        Long bar = 0L;
    }
}

And then decompile them:

$ javap -c Test.class

Compiled from "Test.java"
public class Test {
  public Test();
    Code:
       0: aload_0       
       1: invokespecial #1                  // Method java/lang/Object."<init>":()V
       4: return        

  public static void main(java.lang.String[]);
    Code:
       0: lconst_0      
       1: invokestatic  #2                  // Method java/lang/Long.valueOf:(J)Ljava/lang/Long;
       4: astore_1      
       5: lconst_0      
       6: invokestatic  #2                  // Method java/lang/Long.valueOf:(J)Ljava/lang/Long;
       9: astore_2      
      10: return        
}

I don't see any difference. Use one which is look better for you or corresponds to the conventions.

To verify decompiler I compile this two lines independent and then calculate checksums:

$ sha1sum Test.class_L Test.class_long
292a93b6433b5a451afdb41bd957667c91eebf23  Test.class_L
292a93b6433b5a451afdb41bd957667c91eebf23  Test.class_long

Answer 2

Long foo = (long) 0; // First way of doing it

This way will create an int then cast it to a long.

Long bar = 0L; // Second way of doing it

This is a long literal, so it will only create a long.

I imagine the difference is negligible, but it would be quicker to create a long than to create an int then cast to long.

Edit

As you correctly said, the first way will only allow you to convert an int to a long, which means that in one line, you can only ever create a long the size of an int.. which is pointless and a waste of memory.