Is there a difference between x and x in java

Question

Is there a difference between   x and x   in java

User · Answer

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

User · Answer

Yes  there is a difference  incase of x   postincrement   value of x will be used in the expression and x will be incremented by 1 after the expression has been evaluated  on the other hand   x preincrement   x 1 will be used in the expression   Take an example   public static void main String args          int i   j   k   0      j   k       Value of j is 0     i     j     Value of i becomes 1     k   i       Value of k is 1     System out println k

User · Answer

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

User · Answer

If it s like many other languages you may want to have a simple try   i   0  if  0    i       if true  increment happened after equality check if  2      i     if true  increment happened before equality check   If the above doesn t happen like that  they may be equivalent

User · Answer

x is called preincrement while x   is called postincrement    int x   5  y   5   System out println   x      outputs 6 System out println x      outputs 6  System out println y        outputs 5 System out println y      outputs 6

User · Answer

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

User · Answer

Yes  using   X  X 1 will be used in the expression  Using X    X will be used in the expression and X will only be increased after the expression has been evaluated   So if X   9  using   X  the value 10 will be used  else  the value 9

User · Answer

Yes   int x 5  System out println   x     will print 6 and  int x 5  System out println x       will print 5

User · Answer

The Question is already answered  but allow me to add from my side too    First of all    means increment by one and -- means decrement by one   Now x   means Increment x after this line and   x means Increment x before this line   Check this Example  class Example   public static void main  String args            int x 17 a b        a x          b   x        System out println    x       x     a      a         System out println    x       x      b      b         a   x--        b   --x        System out println    x       x      a      a         System out println    x       x      b      b               It will give the following output   x 19 a 17 x 19 b 19 x 18 a 19 x 17 b 17

User · Answer

OK  I landed here because I recently came across the same issue when checking the classic stack implementation  Just a reminder that this is used in the array based implementation of Stack  which is a bit faster than the linked-list one   Code below  check the push and pop func   public class FixedCapacityStackOfStrings     private String   s    private int N 0     public FixedCapacityStackOfStrings int capacity      s   new String capacity       public boolean isEmpty       return N    0      public void push String item      s N      item       public String pop            String item   s --N       s N    null      return item

User · Answer

These are known as postfix and prefix operators  Both will add 1 to the variable but there is a difference in the result of the statement   int x   0  int y   0  y     x                result  y 1  x 1  int x   0  int y   0  y   x                  result  y 0  x 1

User · Answer

yes    x increments the value of x and then returns x x   returns the value of x and then increments  example   x 0  a   x  b x      after the code is run both a and b will be 1 but x will be 2

User · Answer

In Java there is a difference between x   and   x    x is a prefix form  It increments the variables expression then uses the new value in the expression   For example if used in code   int x   3   int y     x    Using   x in the above is a two step operation    The first operation is to increment x  so x   1   3   4   The second operation is y   x so y   4  System out println y     It will print out  4  System out println x     It will print out  4    x   is a postfix form  The variables value is first used in the expression and then it is incremented after the operation   For example if used in code   int x   3   int y   x      Using x   in the above is a two step operation    The first operation is y   x so y   3   The second operation is to increment x  so x   1   3   4  System out println y     It will print out  3  System out println x     It will print out  4     Hope this is clear  Running and playing with the above code should help your understanding

User · Answer

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

User · Answer

There is a huge difference    As most of the answers have already pointed out the theory  I would like to point out an easy example   int x   1    would print 1 as first statement will x   x and then x will increase int x   x    System out println x     Now let s see   x   int x   1    would print 2 as first statement will increment x and then x will be stored int x     x  System out println x

User · Answer

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   lt init gt     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

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

Examples related to java

Examples related to syntax

Examples related to operators

Examples related to increment