Boxing and Unboxing Java

Question

Integer x1 = 3, x2; // boxing
int y1 = x1; // unboxing
x2 = x1 + (x1 / 2);

Let's suppose we have the following code. Does the 3rd line need 2 unboxings for x1 or we just say its 1 unboxing for x1 and one boxing for the whole result to be assigned to x2?

You can look at the code generated by the Java compiler using the javap tool in the jdk. What does it show you? — Bob Dalgleish, Feb 01 '18 at 19:47
I don't understand what you're trying to tell me. Can you be more specific? I'm new. — Prestyy, Feb 01 '18 at 19:52
It does 2 unboxing, because the two uses of `x1` are handled independently. Once JIT kicks in, and optimizes the code, it may be only one, but you shouldn't rely on that. — Andreas, Feb 01 '18 at 19:54

score 0 · Answer 1 · answered Feb 01 '18 at 19:57

There are IDE plugins, which can be used to demystify boxing and unboxing. This is your code:

   L0
    LINENUMBER 46 L0
    ICONST_3
    INVOKESTATIC java/lang/Integer.valueOf(I)Ljava/lang/Integer;
    ASTORE 1
   L1
    LINENUMBER 47 L1
    ALOAD 1
    INVOKEVIRTUAL java/lang/Integer.intValue()I
    ISTORE 3
   L2
    LINENUMBER 48 L2
    ALOAD 1
    INVOKEVIRTUAL java/lang/Integer.intValue()I
    ALOAD 1
    INVOKEVIRTUAL java/lang/Integer.intValue()I
    ICONST_2
    IDIV
    IADD
    INVOKESTATIC java/lang/Integer.valueOf(I)Ljava/lang/Integer;
    ASTORE 2

Boxing and Unboxing Java

1 Answers1