Is there a difference between ++x and x++ in java?
++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
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.
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: x=1, y=1
int x = 0;
int y = 0;
y = x++; // result: x=1, y=0
Yes,
int x=5;
System.out.println(++x);
will print 6 and
int x=5;
System.out.println(x++);
will print 5.
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.
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
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.
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);
}
}
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.
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
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
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;
}
}
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);
}
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
public static void main(String[] args) {
int a = 1;
int b = a++; // this means b = whatever value a has but, I want to
increment a by 1
System.out.println("a is --> " + a); //2
System.out.println("b is --> " + b); //1
a = 1;
b = ++a; // this means b = a+1
System.out.println("now a is still --> " + a); //2
System.out.println("but b is --> " + b); //2
}
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.
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);
Try to look at it this way:
from left to right do what you encounter first. If you see the x first, then that value is going to be used in evaluating the currently processing expression, if you see the increment (++) first, then add one to the current value of the variable and continue with the evaluation of the expression. Simple
How to change value of a an object of Integer class ex: here I have declared the values of objects x and y to 5 and 6 respectively and I want to swap the value stored in the objects i.e x.intValue() should be 6 and y.intValue() should be 5. Please help.
class second{
public static void main(String[] args) {
System.out.println(In1.a);
Integer x = new Integer(5);
Integer y = new Integer(6);
System.out.println(x.equals(y));
System.out.println(x.intValue());
}
}
Adding a different answer without using a temporary variable
Integer x = 10;
Integer y = 5;
x = x + y;
y = x - y; //y contains 10
x = x - y; //x contains 5
Try,
Integer temp = x;
x = y;
y = temp;
1 ) Have the value of x in a temporary variable
2 ) Assign value of y to x
3 ) Assign value of temp to y
In this way, your values of x and y will get swapped.
You can use a third variable.
public static void main(String[] args){
Integer a = new Integer(6);
Integer b = new Integer(3);
Integer temp = a;
a = b;
b = temp;
}
I have this method in one of my classes:
public int[] getCurrentGridPosition()
{
return new int[]{currentGridPosX, currentGridPosY};
}
Does Java allow something in the likes of:
int x, y;
x = getCurrentGridPosition()[0];
y = getCurrentGridPosition()[1];
If yes, how? If no, why?
Does Java allow something in the likes of:
Yes, there is nothing† wrong with this piece of code. Here a full program with your code to prove it works:
class Main{
private int currentGridPosX = 5,
currentGridPosY = 10;
public static void main(String[] a){
Main m = new Main();
m.test();
}
private void test(){
int x, y;
x = getCurrentGridPosition()[0];
y = getCurrentGridPosition()[1];
System.out.print("x: "+x+"; y: "+y);
}
public int[] getCurrentGridPosition()
{
return new int[]{currentGridPosX, currentGridPosY};
}
}
Try it online.
†: Compile/runtime wise there is nothing wrong. In terms of best practice there are of course things to improve.
If yes, how?
x = getCurrentGridPosition()[0]; will call the method and give an array as result, and will then get the element at index 0, saving it in the field x.
y = getCurrentGridPosition()[1]; will call the method again for a second time and give an array as result, and will then get the element at index 1, saving it in the field y.
So in almost all cases it's best to only call the method once and save the result-array in a variable, and only then access its elements at the indices 0 and 1:
int[] gridPositions = getCurrentGridPosition(); // The method is only called once now
int x = gridPositions[0],
y = gridPositions[1];
System.out.print("x: "+x+"; y: "+y);
Why don't you do it with some more lines of code?
For example, you can access the returned array like this:
int x, y;
int[] myArray = getCurrentGridPosition();
x = myArray[0];
y = myArray[1];
System.out.println("x: " + String.valueOf(x) + ", y: " + String.valueOf(y));
I'm struggling with visualizing why this code would return what it returns.
public class IntObject
{
private int myInt;
public IntObject() { myInt = 0; }
public IntObject(int n) { myInt = n; }
public void increment() { myInt++; }
}
Driver file
public class IntObjectTest
{
public static IntObject someMethod(IntObject obj) {
IntObject ans = obj;
ans.increment();
return ans;
}
public static void main(String[] args) {
IntObject x = new IntObject(2);
IntObject y = new IntObject(7);
IntObject a = y;
x = someMethod(y);
a = someMethod(x);}
In my head, the program is going
a is y
x is y + 1 or a + 1, which in this case is 8
a is x(8) + 1. so then it is 9
When the program ends the values i thought they would be are
x = 8
y = 9
a = 9
but the correct values are all 9. I think I'm having trouble with aliases. Could anyone help me out and explain aliases and how they are working in this code?
Those are not aliases, they are references.
IntObject x = new IntObject(2); // <-- 2
IntObject y = new IntObject(7); // <-- 7
IntObject a = y; // <-- 7
x = someMethod(y); // <-- 8
a = someMethod(x); // <-- 9.
When you call someMethod,
IntObject ans = obj; // ans points to obj.
ans.increment(); // same as obj.increment();
return ans;
I think you expected
IntObject ans = obj.clone(); // ans points to a copy of obj.
ans.increment(); // now it won't modify obj.
return ans;
you are loosing actual reference of x and a.
public static void main(String[] args) {
IntObject x = new IntObject(2); // "new" is creating a memory block. lets call it 0x10
IntObject y = new IntObject(7); // another "new" is creating a memory block. lets call it 0x20
IntObject a = y; // here, a = 0x20
x = someMethod(y); //someMethod actually takes an element and give same element. so here, x shows 0x20
a = someMethod(x); //someMethod actually takes an element and give same element. so here, x shows 0x20, then a shows 0x20
// at the and, y does not change anytime. so all objects addressing to 0x20
}
someMethod(y) returns y, so x = y;
someMethod(x) returns x, so a = x;
==>> y = x = a
I'm trying to prompt the user to enter 3 numbers. After those numbers are entered, I am to add the highest two numbers. The main method is to handle all print statements and is to call the other method. I'm not allowed to use for loop for this problem. The variables from the main, should be passed down to the other method.
I am not sure why I am unable to call the method from the main. Here is my code:
public class HW {
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
System.out.println("Enter three numbers.");
int x = console.nextInt();
int y = console.nextInt();
int z = console.nextInt();
HW.calLargestSum(); //ERROR
HW.calLargestSum(int x, int y, int z); //STILL ERROR
}
public int calLargestSum(int x, int y, int z){
if ( x > y && x > z && y > z )
return x + y;
else if ( y > x && y > z && x > z )
return y + x;
else if ( z > x && z > y && y > x )
return z + y;
return 0;
}
}
You can't call it because you have not instantiated an HW object. Two solutions:
HW hw = new HW();
hw.calLargestSum();
Or make the method static, so that you don't need to instantiate it:
public static int calLargetSum();
Further... ok, so many problems...
HW.calLargestSum(); //ERROR
There is no method calLargestSum(), there is only calLargestSum(int x, int y, int z).
HW.calLargestSum(int x, int y, int z); //STILL ERROR
You need to pass values here. int x is not a value. You need to pass values like:
HW.calLargestSum(1, 2, 3);
Problem
You are making some mistakes when calling the method from main. The non-trivial mistake is that you can't call non-static from static. This happens because if it is not static then it is an instance method. Thus, it requires an instance to access it.
Static Solution
Make your method static. So change your method to:
public static int calLargestSum(int x, int y, int z)
{ ... }
To call the method, you can use:
calLargestSum(1,2,3);
// or in your case.
calLargestSum(x,y,z);
Instance Solution
The other option is to make a new instance of your class (if you don't want it to use static). Like so:
HW hwObj = new HW();
To call use this:
hwObj.calLargestSum(1,2,3);
To See Returned Value/Print
int largest = calLargestSum(x, y, z);
System.out.println(largest);