case constant in switch

[Petra Miller]

Why does this compile? x - 1 is not a constant!

public class Switch3 {
	public static void main(String[] args) {
		final short x = 2;
		switch (1) {
		case x - 1:
			System.out.print("1 ");
		}
	}
}

[Campbell Ritchie]

I am going to assume your having three copies of this thread simply means you pushed the same button thrice.

Apart from the fact that shorts should only be worn above the knee, and not used in normal programming:-

But x − 1 is a constant. You are declaring a compile‑time constant 2 and assigning it to a final local variable x, so x also becomes a compile‑time constant. Because line 5 only contains expressions using ordinary arithmetic operators on compile‑time constants, it is a compile‑time constant itself. You can see that from the bytecode, using the javap tool:-javac Switch3.java [critchie@...]$ javap -c Switch3 Compiled from "Switch3.java" public class Switch3 {  public Switch3();    Code:       0: aload_0       1: invokespecial #1                  // Method java/lang/Object."<init>":()V       4: return  public static void main(java.lang.String[]);    Code:       0: iconst_1       1: lookupswitch  { // 1                     1: 20               default: 28          }      20: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;      23: ldc           #3                  // String 1      25: invokevirtual #4                  // Method java/io/PrintStream.print:(Ljava/lang/String;)V      28: return }Line 13 and 15 how you are using 1 as a constant.

[Petra Miller]

Thank you for you answer!

[Campbell Ritchie]

Petra Miller wrote:Thank you . . .

That's a pleasure