Expressions and operators

Question

Hello
quick question
is
count--;equivalent to
count = count - 1;meaning they produce the same output...

Answer 1

corlettk wrote:
That's it... I stick to my rule.... NEVER, never, never decrement and assign in one statement.... It's just too confusing for a moron like me.You can even call decrement on Double (or other numeric wrapper types):
Double a = 1.0;
a++;
++a;
System.out.println(a);The above prints 3.0
Or, try this:
import java.util.List;
import java.util.ArrayList;
public class Test {
   public static void main (String [] args)
      Integer a = 1;
      Integer b = a;
      a++;
      a--;
      System.out.println("a and b:");
      printOutput(a, b);
      System.out.println();
      Integer c = 1200;
      Integer d = c;
      c++;
      c--;
      System.out.println("c and d:");
      printOutput(c, d);
      System.out.println();
      List<Integer> list = new ArrayList<Integer>();
      list.add(d);
      d++;
      System.out.println("List: " + list);
      System.out.println("d: " + d);
      System.out.println("list.get(0) == d: " + (list.get(0) == d));
   public static void printOutput(Integer left, Integer right)
      System.out.println("left, right: " + left + ", " + right);
      System.out.println("left == right: " + (left == right));
      System.out.println("left.equals(right): " + left.equals(right));
}Output:
a and b:
left, right: 1, 1
left == right: true
left.equals(right): true
c and d:
left, right: 1200, 1200
left == right: false
left.equals(right): true
List: [1200]
d: 1201
list.get(0) == d: falsea and b are == because of caching the values of certain Integers.
c and d are not == because they aren't cached.
You put d into the list, but you don't get the same value back out, because the ++ assigns a new Integer instance to the reference d. That makes sense, as you wouldn't want a mutable Integer. Calling d++ doesn't affect the Integer instance that was put into the list, which also makes sense. It's the same as setting d to a new Integer explicitly:
d = Integer.valueOf(5);This assignment obviously wouldn't affect the Integer in the list.