Please explain what is the difference between ++x and x++?

Question

Please explain what is the difference between ++x and x++?

Answer 1

There is no difference when used alone, the difference is when you use them in an expression.

a++ evaluates a, then increments it (post-incrementation).
++a increments a, then evaluates it (pre-incrementation).

Example:
int a = 1;
int b = a++; //b = 1, a = 2
System.out.println(b); //prints 1
a = 1;
b = ++a; //b = 2, a = 2
System.out.println(b); //prints 2

Answer 2

If int x=5; and and y=5;

System.out.println(x++); //5
System.out.println(x); //6

System.out.println(++y); //6

hope it helps

Answer 3

I'm probably going to say the same thing Lara said..
++x just increments x before working with it and,
x++ increments x after working with it..
there isnt much difference until used in a function..

Answer 4

++x is prefix. x++ is postfix.

In the case of y =++x , y will become (x+1), x will become (x+1). Which means that prefix adds the value before processing the data/command.

However, in the case of y = x++ , y will become x and x will become (x+1).
Which also means that postfix process the data/command first before it adds the value

Answer 5

Now x++ means Increment x after this line and ++x means Increment x before this line. With i++, it's called post increment, and the value is used in whatever context then incremented; ++i is pre-increment increments the value first and then uses it in context

for example:
#include <iostream>
using namespace std;

main() {
int a = 21;
int c ;

// Value of a will not be increased before assignment.
c = a++;
cout << "Line 1 - Value of a++ is :" << c << endl ;

// After expression value of a is increased
cout << "Line 2 - Value of a is :" << a << endl ;

// Value of a will be increased before assignment.
c = ++a;
cout << "Line 3 - Value of ++a is :" << c << endl ;
return 0;
}

answer for this is:
Line 1 - Value of a++ is :21
Line 2 - Value of a is :22
Line 3 - Value of ++a is :23

There isn't much difference until used in a function.

I hope it helps!!! Thank You

Answer 6

++ is a increment operator.
there is two types of increment operators postfix (var++) & prefix (++var).

as the name suggests PREFIX OPERATORS EVALUATE BEFORE. & POSTFIX OPERATORS EVALUATE AFTER.

let see some basic example like

int a=10;
int b=a++
/*old value of a will be assigned to b , the value of a increased by 1 means a=11 & b=10 after this statement.*/

int a=10;
int b=++a;
/*new value of a will be assigned to b after increasing value of a by 1 means a=11 & b=10*/

NOTE:

now here one more thing you should know about prefix & post fix is IN JAVA POSTFIX HAVE HIGHEST PRECEDENCE. yes I have not make any mistake in typing, postfix will evaluate before any other operators.

let see one interesting example...

int a=10;
a=a++;

think what should be the value of a...
let see the code carefully.
first of all there is assignment operator (=) is there.
so it will evaluate first as any equation in Java evaluated lest to right. but in the case of '=' the right side of operator evaluated first so a++ IS EVALUATED FIRST. now at the time of evaluation the VALUE OF A WILL BE INCREASED & STORE IN MEMORY OF VARIABLE A. BUT THE OLD VALUE OF A WILL BE USED TO EVALUATE EXPRESSION. so when assignment operator executed the value of a actually incremented in memory but the old value is in buffer or temporary memory which is used to evaluate expression so final 10 is assigned to a.

do not make mistake in understanding of postfix & prefix. in old gcc compiler for c language the evaluation of the postfix done differently. for example you can get different value of c in different c compiler in following example...

int a=10;
int b=13;
int c;
c= a++ + ++b + --a + --b + a;

in old gcc compiler you may get c = 53 & in new gcc compiler you will get c = 57.

so do not make mistake. enjoy programming.

Answer 7

Hello, PHP explains so:

$a=2; $b=$a++ // $a=3, $b=2
$a=2; $b=++$a // $a=3, $b=3

Answer 8

for example
y=4
z=++y
so both y and z = 5

but

y=4
z=y++
y=5 and z=4

Answer 9

please solve my problem
int x =5;
int y = x++;
//out put is 5. why?
please give me answer with explain

Answer 10

Since x++ is post increment so it first assign value of x to y the. Increment value of x by 1.

Answer 11

for example
y=4
z=++y
so both y and z = 5

but

y=4
z=y++
y=5 and z=4

Answer 12

prefix inc
{cout<<++a;} equal {a=a+1;cout<<a;}
postfix inc
{cout<<a++;} equal {cout<<a;a=a+1;}

Answer 13

a++ use the value of 'a' first and then increment it.
But,++a increment the value by 1 and use the value of 'a'.

Answer 14

processing

Answer 15

++ is a increment operator.
there is two types of increment operators postfix (var++) & prefix (++var).

as the name suggests PREFIX OPERATORS EVALUATE BEFORE. & POSTFIX OPERATORS EVALUATE AFTER.

Answer 16

++x is prefix so during execution the value of x increased by one during execution of line whereas the x++ is a post fix so it results after the execution of the line.

Answer 17

this is a year old why are yall still commenting lmao, theres already 100 other answers

Answer 18

x++ is post-increament and ++x is pre-increament
let see
int a=0;
System.out.println(++a);//output will be 1
but
a=0;
System.out.println(a++);//output will be 0 in this case but value of a is equal to 1 now.

Answer 19

x=1;
y = x++; post increment
-----------------------------------------------------
first: y = x = 1
second x++ : x = 2

x=1;
y = ++x; pre increment;
-----------------------------------------------------
first ++x: x = 2
second: y = 2;

Answer 20

you will know the diffrent when seeing this example:
for example we have a=1 and i=3
a=i++ that mean a=3 and i=4 but for a=++i we have a=4 and i=4