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));
As I have read online that Java is pass by value and a general swap function won't swap the two values. I also read that it's not possible to swap the values of primitive types. I am wondering why the following program works and displays different valies after swap ?
public class swapMe {
public static void main(String[] args) {
int x = 10 , y = 20;
System.out.println("Before");
System.out.println("First number = " + x);
System.out.println("Second number = " + y);
int temp = x;
x = y;
y = temp;
System.out.println("After");
System.out.println("First number = " + x);
System.out.println("Second number = " + y);
}
}
Is it like somewhere, the original values of x = 10 and y = 20 are still stored somewhere and the swapped values displayed are not correct? Please advise. Thanks
Not entirely sure where you're getting that information, but let's take this one at a time.
As I have read online that Java is pass by value and a general swap function won't swap the two values.
Correct...if the expectation of the swap is to happen by virtue of calling a method.
public void swap(int x, int y) {
int tmp = x;
x = y;
y = tmp;
}
// meanwhile, in main
int x = 10;
int y = 20;
swap(x, y);
System.out.println(x); // still prints 10
System.out.println(y); // still prints 20
Incorrect...if the swap happens inside the method and is utilized somehow.
public void swap(int x, int y) {
int tmp = x;
x = y;
y = tmp;
System.out.println(x); // will print 20 from main
System.out.println(y); // will print 10 from main
}
// meanwhile, in main
int x = 10;
int y = 20;
swap(x, y);
System.out.println(x); // still prints 10
System.out.println(y); // still prints 20
I also read that it's not possible to swap the values of primitive types.
No, this is perfectly possible to do. You can always reassign variables.
As to why your example above works, it's because you're holding onto one of the values while you reassign one of its variables. You're basically putting one value to the side while you copy it over.
Slowly...
int x = 10;
int y = 20;
int tmp = x; // tmp = 10
x = y; // x = 20, tmp = 10
y = tmp; x = 20, y = 10; tmp = 10 (but that doesn't matter)
When I execute this code the output is "140" which is "28*5" but it should be "150" which is "28+31+30+31+30" it should calculate the days between 2 months "feb" and "july"... So it means that the for loop isn't working correctly or what ? and why is that ! can you help me here ??
PS: I tried to change the j++ in the loop to j+1 but Android Studio Says"that's not a statement"
int[] pair = {1,3,5,7,8,10,12};
int[] impair = {4,6,9,11};
int x=0;
int j;
int year=2015;
int mm=2;
int month=7;
String msg="";
if (month>mm) {
for (j = mm; j<month; j++){
if (Arrays.asList(impair).contains(j)){
x = 31 + x;
}else if(Arrays.asList(pair).contains(j)){
x = 30 + x;
}else{
if (year%4==0) {
x= 29 + x;
}else{
x= 28 + x;
}
}
}
System.output.println(x);
}
You are attempting to convert the int[] to a List<Integer> by calling Arrays.asList. But that results in a List<int[]> of a single element (the original int[]), which doesn't contain any value of j. The reason is given in Arrays.asList() not working as it should? -- it's a generic method, and the type parameter must be a reference type. int[] is a reference type (as all arrays are), but int isn't.
That is why the tests all fail and 28 is repeatedly chosen to be added.
Change the declared types of pair and impair from int[] to Integer[], so that Arrays.asList will infer the type as Integer (a reference type) correctly. Then the contains method will work as expected. With this change I get 150.
This is because the asList() requires a class object which is either a collection and Iterable. You could modify your code in the following way:-
Integer[] pair = {1,3,5,7,8,10,12};
Integer[] impair = {4,6,9,11};
int x=0;
int j;
int year=2015;
int mm=2;
int month=7;
String msg="";
if (month>mm) {
for (j = mm; j<month; j++){
if (Arrays.asList(impair).contains(new Integer(j))){
x = 31 + x;
}else if(Arrays.asList(pair).contains(new Integer(j))){
x = 30 + x;
}else{
if (year%4==0) {
x= 29 + x;
}else{
x= 28 + x;
}
}
}
System.output.println(x);
}
This should give you the correct output.
This is because your loop never enters the first two if blocks as that is not how aslist(array).contains(element) works
Java, Simplified check if int array contains int
Checking whether an element exist in an array