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
Related
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)
class Output {
public static void main(String [] args) {
Output o = new Output();
o.go();
}
void go() {
int y = 7;
for(int x = 1; x < 8; x++) {
y++;
if (x > 4) {
System.out.print(++y + " ");
}
if (y > 14) {
System.out.println(" x = " + x);
break;
}
}
}
}
Your doubts probably are related to the use of:
x++: which increments the value after use of current
++y: which first increments and then use the new value
Running it we have these variables states in paper
Click here to see the output
Note when x=5 and x=6 , the ++y is executed when printed, and by this you see two increments for each iteration.
Hope this clarify your doubt.
Luigi.
This question already has answers here:
What is x after "x = x++"?
(18 answers)
Closed 8 years ago.
I was executing the following code
class First
{
public static void main(String arg[])
{
char x= 'A';
x = x++;
System.out.println(x);
}
}
Here the output is A.
My question is why didn't x get incremented before printing.
You're using the post-increment operator incorrectly - you don't need to use assignment as well. And in this case it undermines what you're trying to do.
For context, remember that the post-increment operator increases the value, and returns the old value. That is, x++ is roughly equivalent to:
int x_initial = x;
x = x + 1;
return x_initial;
Hopefully now you can see why your code is failing to change x. If you expand it, it looks like:
char x= 'A';
char y;
{
y = x;
x = x + 1;
}
x = y;
System.out.println(x);
and the net effect of the assignment, is to set x back to what it was originally.
To fix - you can simply call x++. Or, if you want to make it clear there's some sort of assignment happening, x += 1 or even just x = x + 1 would do the same thing.
class First
{
public static void main(String arg[])
{
char x= 'A';
x = x++; // it is post increment as ++ sign is after x
System.out.println(x);
}
}
Post Increment(x++) : First execute the statement then increase the value by one.
Pre Increment (++x) : First increase the value by one then execute the statement.
According to this answer https://stackoverflow.com/a/12020435/562222 , assigning an object to another is just copying references, but let's see this code snippet:
public class TestJava {
public static void main(String[] args) throws IOException {
{
Integer x;
Integer y = 223432;
x = y;
x += 23;
System.out.println(x);
System.out.println(y);
}
{
Integer[] x;
Integer[] y = {1,2, 3, 4, 5};
x = y;
x[0] += 10;
printArray(x);
printArray(y);
}
}
public static <T> void printArray(T[] inputArray) {
for(T element : inputArray) {
System.out.printf("%s ", element);
}
System.out.println();
}
}
Running it gives:
223455
223432
11 2 3 4 5
11 2 3 4 5
The behavior is consistent. This line:
x += 23;
actually assigns a different Integer object to x; it does not modify the value represented by x before that statement (which was, in fact, identical to object y). Behind the scenes, the compiler is unboxing x and then boxing the result of adding 23, as if the code were written:
x = Integer.valueOf(x.intValue() + 23);
You can see exactly this if you examine the bytecode that is generated when you compile (just run javap -c TestJava after compiling).
What's going on in the second piece is that this line:
x[0] += 10;
also assigns a new object to x[0]. But since x and y refer to the same array, this also changes y[0] to be the new object.
Integer is immutable, so you cannot modify its state once created. When you do this:
Integer x;
Integer y = 223432;
x = y;
x += 23;
The line x += 23 is assigning a new Integer value to x variable.
Arrays, on the other hand, are mutable, so when you change the state of the array e.g. changing one of the elements in the array, the other is affected as well.
When you do this x += 23;, you actually create another object, and you made x point on this new object. Therefore, x and y are 2 different objects.
This question already has answers here:
Is there a difference between x++ and ++x in java?
(18 answers)
Closed 10 years ago.
What is the difference between x++ and ++x in Java
Can anybody please tell me the difference of the above by refering the below code,
class Example{
public static void main(String args[]){
int x=10;
int y;
y=x++; //Prints 11 10
System.out.println(x+"\t"+y)
}
}
class Example{
public static void main(String args[]){
int x=10;
int y;
y=++x; //Prints 11 11
System.out.println(x+"\t"+y)
}
}
y=x++ assigns x to y, then increments x.
y=++x increments x, and then assigns it to y.
++x is the pre-increment. i.e., The value of x is first incremented and then assigned to x.
x++ is post increment. i.e., the value of x is assigned first and then incremented.
y=x++;
is essentially same as
y =x;
x= x+1;
y=++x; is same as
y= (x+1);
Pre and Post increment. Increment BEFORE assignment and increment AFTER assignment respectively.
The difference is that in the first case (x++) Java first resolves the assignment issue and then increments x. In the other case (++x) Java first resolves the increment and then the assignment.
In the following code you will see the difference:
#Test
public void test1() {
int x = 1;
int y = 1;
y = 2 + x++;
assertEquals(2, x);
assertEquals(3, y);
}
#Test
public void test2() {
int x = 1;
int y = 1;
y = 2 + ++x;
assertEquals(2, x);
assertEquals(4, y);
}
As you can see, x always will be incremented, but the difference is the order in which the expression is resolved.
Hope it would be useful!