Why is Scala's == different in case of Int?

Question

Isn't it supposed to call equals behind the scene?

scala> 1 equals 1l
res2: Boolean = false

scala> 1 == 1l
res3: Boolean = true

Answer 1

as @MichaelLang suggests, this answer can help you, concretely:

comparing two primitives (boxed or unboxed) with == should always give the result you would have gotten by comparing those values as unboxed primitives. When you call equals directly, you are skipping all that softening logic and instead treated to java's theory that two boxed values of different types are always unequal.

If you analyze the bytecode:

1 == 1l produces

         0: aload_0       
         1: invokespecial #19                 // Method java/lang/Object."<init>":()V
         4: aload_0       
         5: putstatic     #21                 // Field MODULE$:L;
         8: aload_0       
         9: iconst_1      
        10: putfield      #17                 // Field res0:Z
        13: return        

directly compares the values of both primitive 1's, but if you examine:

1 equals 1l then

     0: aload_0       
     1: invokespecial #19                 // Method java/lang/Object."<init>":()V
     4: aload_0       
     5: putstatic     #21                 // Field MODULE$:L;
     8: aload_0       
     9: iconst_1      
    10: invokestatic  #27                 // Method scala/runtime/BoxesRunTime.boxToInteger:(I)Ljava/lang/Integer;
    13: lconst_1      
    14: invokestatic  #31                 // Method scala/runtime/BoxesRunTime.boxToLong:(J)Ljava/lang/Long;
    17: invokevirtual #35                 // Method java/lang/Object.equals:(Ljava/lang/Object;)Z
    20: putfield      #17                 // Field res0:Z
    23: return      

Where you can see that is boxing 1l primitive into a Long object and then performing the equals.