Is there a difference between x++ and ++x in java?

Question

Is there a difference between ++x and x++ in java?

Answer 1

Yes, the value returned is the value after and before the incrementation, respectively.

class Foo {
    public static void main(String args[]) {
        int x = 1;
        int a = x++;
        System.out.println("a is now " + a);
        x = 1;
        a = ++x;
        System.out.println("a is now " + a);
    }
}

$ java Foo
a is now 1
a is now 2

Answer 2

I landed here from one of its recent dup's, and though this question is more than answered, I couldn't help decompiling the code and adding "yet another answer" :-)

To be accurate (and probably, a bit pedantic),

int y = 2;
y = y++;

is compiled into:

int y = 2;
int tmp = y;
y = y+1;
y = tmp;

If you javac this Y.java class:

public class Y {
    public static void main(String []args) {
        int y = 2;
        y = y++;
    }
}

and javap -c Y, you get the following jvm code (I have allowed me to comment the main method with the help of the Java Virtual Machine Specification):

public class Y extends java.lang.Object{
public Y();
  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_2 // Push int constant `2` onto the operand stack. 

   1:   istore_1 // Pop the value on top of the operand stack (`2`) and set the
                 // value of the local variable at index `1` (`y`) to this value.

   2:   iload_1  // Push the value (`2`) of the local variable at index `1` (`y`)
                 // onto the operand stack

   3:   iinc  1, 1 // Sign-extend the constant value `1` to an int, and increment
                   // by this amount the local variable at index `1` (`y`)

   6:   istore_1 // Pop the value on top of the operand stack (`2`) and set the
                 // value of the local variable at index `1` (`y`) to this value.
   7:   return

}

Thus, we finally have:

0,1: y=2
2: tmp=y
3: y=y+1
6: y=tmp

Answer 3

With i++, it's called postincrement, and the value is used in whatever context then incremented; ++i is preincrement increments the value first and then uses it in context.

If you're not using it in any context, it doesn't matter what you use, but postincrement is used by convention.

Answer 4

When considering what the computer actually does...

++x: load x from memory, increment, use, store back to memory.

x++: load x from memory, use, increment, store back to memory.

Consider: a = 0 x = f(a++) y = f(++a)

where function f(p) returns p + 1

x will be 1 (or 2)

y will be 2 (or 1)

And therein lies the problem. Did the author of the compiler pass the parameter after retrieval, after use, or after storage.

Generally, just use x = x + 1. It's way simpler.

Answer 5

Yes.

public class IncrementTest extends TestCase {

    public void testPreIncrement() throws Exception {
        int i = 0;
        int j = i++;
        assertEquals(0, j);
        assertEquals(1, i);
    }

    public void testPostIncrement() throws Exception {
        int i = 0;
        int j = ++i;
        assertEquals(1, j);
        assertEquals(1, i);
    }
}

Answer 6

Yes,

int x=5;
System.out.println(++x);

will print 6 and

int x=5;
System.out.println(x++);

will print 5.