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
Why is the output 25?
// CODE 1
public class YourClassNameHere {
public static void main(String[] args) {
int x = 8;
System.out.print(x + x++ + x);
}
}
Hi!
I am aware that the above code will print 25. However, I would like to clarify on how x++ will make the statement be 8 + 9 + 8 = 25.
If we were to print x++ only as such, 8 will be printed while x will be 9 in-memory due to post incrementation.
// CODE 2
public class YourClassNameHere {
public static void main(String[] args) {
int x = 8;
System.out.print(x++);
}
}
But why is it that in code 1 it becomes 9 ultimately?
I thank you in advance for your time and explanation!
Here is a good way to test the reason that equals to 25 is because the third x is equal to 9.
public class Main {
public static void main(String[] args) {
int x = 8;
System.out.println(printPassThrough(x, "first") + printPassThrough(x++, "second") + printPassThrough(x, "third"));
}
private static int printPassThrough(int x, String name) {
System.out.println(x + " => " + name);
return x;
}
}
Result
8 => first
8 => second
9 => third
25
I think it's worth of clarify:
x++ -> operate x , and then increment x (meaning x=x+1)
++x -> increment x (meaning x=x+1) , and then operate x
x-- -> operate x , and then decrement x (meaning x=x-1)
--x -> decrement x (meaning x=x-1) , and then operate x
I have been struggling with an exercise in the Java Headfirst book( CH5: p121 for reference). It's a loop inside another loop which adds/substracts some values from instance variables.
Input:
x = x + 3
Outputs:
x= 54 y = 6
public class MixFor5 {
public static void main(String[] args) {
int x = 0;
int y = 30;
for (int outer = 0; outer < 3; outer++) {
for (int inner = 4; inner > 1; inner--) {
x = x + 3;
y = y - 2;
if (x == 6) {
break;
}
x = x + 3;
}
y = y - 2;
}
System.out.println(x + " " + y);
}
}
My result is when doing it by myself with a notepad is x=42 y = 8 because then both loop conditions are met. What am i doing wrong? where did I go wrong in my thoughtprocess?
these are my notes -> pastebin note
I have not tried debugging first because I want to figure this by myself first so that I don't make the same mistakes in the future.
Thanks in advance,
tvanderv
if(x == 6) will never get true. The reason behind this is,
When inner = 4
x = x + 3 executes two times i.e. means x = 6.
then, inner = 3
now first x = x + 3 (before if(x == 3) condition) will give output x = 9. So x > 6 it will not break loop.
You did this step wrong in your notes.
It's been 3 days since I start to learn Java.
I have this program and I don't understand code in main method with ++ and -- operators. I don't even know what to call them(name of these operators)
Can anyone explain me what's all about.
class Example {
public static void main(String[] args) {
x=0;
x++;
System.out.println(x);
y=1;
y--;
System.out.println(y);
z=3;
++z;
System.out.println(z);
}
}
These are called Pre and Post Increment / Decrement Operators.
x++;
is the same as x = x + 1;
x--;
is the same as x = x - 1;
Putting the operator before the variable ++x; means, first increment x by 1, and then use this new value of x
int x = 0;
int z = ++x; // produce x is 1, z is 1
int x = 0;
int z = x++; // produce x is 1, but z is 0 ,
//z gets the value of x and then x is incremented.
++ and -- are called increment and decrement operators.
They are shortcuts for writing x = x+1 (x+=1) / x = x-1 (x-=1). (assumed that x is a numeric variable)
In rare cases you could worry about the precedence of the incrementation/decrementation and the value the expression returns: Writing ++x it means "increment first, then return", whereas x++ means "return first, then increment". Here we can distinguish between pre- and post increment/decrement operators.
Please someone explain the ("x = " + x) part of the code.
public class While-With-Nested-If {
public static void main(String [] args) {
int x = 1;
while(x < 100) {
System.out.println("x = " + x);
if(x % 2 == 0) {
x++;
} else {
x *= 2;
}
}
}
}
In this case the operator is used to concatenate a string with the string representation of x.
It depends on the operand type.
For String type operands it creates a new String instance (String objects are immutable) and assigns to it the concatenation of two operands.
For numeric types it works as addition operator.
In this case, it concatenates a String with the string representation of x. For example: x = 42;
The "+" operator acts as syntactic sugar for the concatenation operator in regards to String operations.
It is used to concatenate the two strings in your case.
When you write
String a = b + c + d;
then it gets converted into:
String a = new StringBuilder(b).append(c).append(d).toString();
You may refer Oracle docs for more details
In this code block while loop will iterate till x is less than 100 in order to print all the values of x during "while" loop executes the System.out.println("x = " + x); is used.
Here java will send each value of x to output console by appending it to the "x = " string(text) so on each iteration of while loop you will get output on console like
x = 1
x = 2
x = 3
and so on...
x = 99
The + operator concatenates if the operands' type is string, and it performs summing if the operands are ints, floats or doubles.
Here is the out put:
X = 1, X = 2, X = 3, X = 6, X = 7, X = 14, X = 15, X = 30, X = 31, X = 62, X = 63...
int x = 10;
x += x++;
System.out.println(x);
why the answer of above statement is 20 ?
The operator += is an addition assignment operator. Like Alya said above, x += x++ is equivalent to x = x + x++, which in your case is x = 10 + 10. However, it's a very messy statement and I'll explain why towards the end of this post.
Now, you're probably thinking "Why is it 20 and not 21 (10 + 11) since you have the ++?" and that's valid. There's actually a difference between a post-increment and a pre-increment. x++ is the post-increment and will actually evaluate the value of x first and THEN increment x, while ++x is the pre-increment which will increment x and THEN evaluate the value of x.
For example, x = 10; System.out.println(x++); System.out.println(x); will print 10 and then print 11 because the first print line prints x and THEN performs the ++ calculation, making x 11 which the next line prints. Conversely, x = 10; System.out.println(++x); System.out.println(x); will print 11 on both print statements.
Going back to why I said x += x++; is very messy is because technically the ++ operator isn't performed in this case. x++ is technically the same as x=x+1 and remembering that x+=y is the same as x = x+y) , the line x += x++; is kind of like saying x = x + (x = x + 1); which is kind of weird looking because you do 2 assignment statements in one and won't actually "work how you want it". Back to your example int x = 10; x += x++; if you print x, you will get 20 even though you could look at it as: x is now the value of x + the value of x, then finally + 1 to it. But unfortunately, that's not how it works.
To solve your problem, if you change your code from a post-increment to a pre-increment, then it should work, ie: x+=++x; will print your 11 but I would argue the that's quite unreadable and a bit confusing. x+=x; x++; System.out.println(x); is easier to follow.
x++ will execute first. It returns x and then increments x by 1.
Finally, the += operator will add to x the return value of x++, which was 10.
Thus, x will be 20 and it will overwrite the changes to x by the statement x++.
So first x is initialized to be 10. Then the x++ has higher precedence so that gets carried out first. the "++" is a post-increment in this case (because it is after the variable as opposed to pre-increment which would be ++x). Post-increment means "first use the variable then increment it by one" so in this case it first uses x to be 10 then increments it to 11 after it is used. Then we look at the "+=" which is short hand for "x = x+x++". so we have x = 10+10 which = 20. If you were to carry this out again it would equal x = 20+20 = 40.
In this particular case, the x++ isn't necessary as x is reassigned the value after it is incremented each time.
int x = 10; x += x++;
will equal to x=x+x
where x++ mean use the x value then increament it , so it's value will be 10
so the result will equal 20
if you want to see the change of the x , see this example:
int x = 10;
int y = 10;
y +=x++;
System.out.println(y);
System.out.println(x);
will print :
y=20
x=11////////////according to x++ and without to overwrite it
//
// Shows how increments work.
//
int i = 0;
System.out.println(i);
i++; // Add one
System.out.println(i);
i += 2; // Add two
System.out.println(i);
i += 3; // Add three
System.out.println(i);
++i; // Add one
System.out.println(i);
i += i; // Added itself
System.out.println(i);
//
// Uses increments and assigns.
//
int v = 0;
v = i++; // Increment after value copy
System.out.println(v);
System.out.println(i);
v = ++i; // Increment before value copy
System.out.println(v);
System.out.println(i);
//Output
0 -
1
3
6
7
14
14
15
16
16
x+=x++ first assigns the value to x and then increments (post-increment)
x+=++x first increments then assign the value to x (pre increment)
there are two types of increments/decrements in programming
1. pre-increment/decrement
2. post-increment/decrement
In programming both of these have same operations but differ in there nature as they both used for increment or decrement; they can be written as,
x+=1; (increment by 1)
x-=1; (decrement by 1)
you can use a variable instead in the above cases as well