++ operator in java [duplicate]

++ operator in java [duplicate] - java

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Closed 10 years ago.
Possible Duplicate:
java operator ++ problem
This is the code. I know the difference between c++ and ++c.
public class sample {
public static void main(String[] b){
int count = 0,a=0;
for (int i = 0; i < 3; i++){
count=count++;
System.out.println(count);
}
But what I am expecting is
count=count;count=count+1;//output has to be 1 2 3
but the output is 0 0 0.

Your prediction is wrong.
count++ will increment count by 1 and return the old value (0). Which is your case. Afterwards, you assign the old value (0) to your count variable. To make it more understandable, just look at this code
count = count; // is the same as count = count++;
Don't use count = count++;, just use count++

In Java, this code is guaranteed to keep the variable with the same value.
It's like:
int temp;
temp = count;
count = count +1;
count = temp;
to acheieve what you want, write:
count++; //or
count += 1; //or
count = count +1;

Replace count=count++; with count++; .

count = count ++;
Here's what is happening.
First, count++ is evaluated, which evaluated to 0, but increments count. And this 0 is assigned to count. So count remains 0. The following is different, because ++count evaluates to 1, 2...
count = ++count;

I think this will give you a pretty good explanation.
Consider this class:
public class T
{
public void f() {
int count = 0;
count = count++;
}
}
This is the associated byte code:
public void f();
Code:
0: iconst_0
1: istore_1
2: iload_1
3: iinc 1, 1
6: istore_1
7: return
}
iconst_0 loads the constant 0 onto the stack (this is for assigning the variable count with value 0
istore_1 stores stack value (0 right now) into variable 1
iload_1 loads int value from variable 1 (0 right now) onto the stack
zinc 1, 1 increments by 1 variable 1 (count = 1 right now)
istore_1 stores stack value (0 right now from step #3) into variable 1
return
Now it should be pretty clear how count = count++ gets compiled in Java.

I slightly modified your code and I have made it to work
public class sample {
public static void main(String[] b){
int count = 0,a=0;
for (int i = 0; i < 3; i++){
count++;
System.out.println(count);
}
}
}
You don't have to reassign the value of count++ to count. Java will do it for you. I added some parentheses which were missing in your code. I hope this helps.

Related

Facing problem with the output of this code

My friend gave me this code and i cant seem to find the error in it. I am attaching the code below:
import java.util.*;
public class prg {
public static void main(String[] args) {
int n;
int count;
int a=0,b=1;
int c=0;
Scanner kb=new Scanner(System.in);
n=kb.nextInt();
int ar[]=new int[100];
ar[1] = 2;
for(int i=3;i<=n;i+=2)
{
count=0;
for (int j = 2; j < i ;j++ ) {
if (i % j == 0) {
count++;
break;
}
}
if(count==0)
ar[i]=i;
}
for(int i=0;i<=n;i+=2)
{
a = b;
b = c;
c = a + b;
ar[i]=c;
}
for(int i=0;i<14;i++)
System.out.print(ar[i]+" ");
}
}
So, the even index is storing the fibonacci series and the odd index is storing prime number.
Problem: the rest of the code is working fine but the 9th index of 'ar' array is printing 0 i dont know why and because of it the output is showing wrong.
Take input n as 14 and check the code please.
Thankyou in advance.
PS: i have solved this question in one other way so i request you to not give answers like 'try my method, its not efficient'. I just want to know what is going wrong at INDEX 9 of the array.
Edited: facing problem with the Prime Number loop.

When i is 9, your code correctly identifies that it is not a prime number, so count is not 0. This causes this line to not run:
ar[i]=i;
And then you increase i by 2 to check the next odd number. This means that you never set index 9 of the array to anything, so it remains at its default value - 0.
To fix this, you should introduce a new variable possiblePrime to keep track of which number you are checking. Increase this variable every iteration of the outer for loop, and increase i only when possiblePRime is prime. Also, change the above line to:
ar[i] = possiblePrime;

9 is not a prime number, so it sets nothing in the array. 0 is the default value so it gets printed.

Method is being called 8 times and I do not understand why?

My test review is asking "How many times will niceHippo () be called?" and the correct answer is 8. I'm having trouble understanding this as no matter how I look at it I'm not seeing how it results in 8. Please help
public class Animals{
public static String niceHippo()
{
String hippo = "Nice Hippo";
return hippo;
}
public static String niceLion(){
String lion = "Nice Lion";
return lion;
}
public static void main(String[] args){
int count = 13;
String stringOut = "I love this class ";
do
{
stringOut = "Animals can be messy ";
for (int order = 1; order < 5; ++ order)
for (int copy = 1; copy <= 2; copy++)
System.out.println(niceHippo());
System.out.println(niceLion());
}while (count != 13);
count = 13;
while (count > 10)
{
count--;
}
System.out.println(stringOut + count);
}
}

In your code you are iterating outer loop for (int order = 1; order < 5; ++ order) for order = 1 to 4 and for each iteration copy value for (int copy = 1; copy <= 2; copy++) will be 1 and 2 and niceHippo is called from inner loop, so as a result niceHippo is being called 8 times

For the loop using copy, each time that's ran you print nice hippo twice because copy <=2
Then the loop using order is going to ask that copy loop to run 4 times because order <5
so in the end nice hippo prints 8 times because one loop prints nice hippo twice and its asked to do that 4 times by the outer loop.

Find value that occurs consecutively most nunber of times in array

The following code is in java lang:
int[] array = {1,2,2,3,3,3,2,2,1};
int k = 2;
What i need to do is find the value (in this case 3), that occurs consecutively more than k times. There can be only one value that occurs>k and if no such value exists, print -1
Explanation: 1 occurs only 1 time consecutively. 2 occurs 2 times but its not>k. 3 occurs 3 times, which is>k. Since, there can be only one possible answer, you can stop searching for answer in further values and print 3.
The time limit for code is 0.25s
Update : What I've tried so far
import java.util.*;
import java.lang.*;
import java.io.*;
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
int[] array = {1,2,2,3,3,3,2,2,1};
int k=2;
int result = -1;
for(int lom =0;lom<array.length-1;lom++){
int temp = 0;
int lo=lom;
while(array[lo]==array[lom]){
temp+=1;
if(lo<array.length){lo++;}else{break;}
}
if(temp>k){
result = array[lom];
break;
}
}
System.out.println(result);
}
}
When I try to solve about 10 queries through this, it takes up 2 seconds. I got to complete it in 1. Can you suggest some method to optimize the code, so that I can research upon it and then try again.

You need not to traverse twice with while loop and lo
1.) Traverse using loop
2.) If keep a track of element sequence
3.) if element changes , reset track
4.) when the track reaches above k break the loop
int[] array = {1,2,2,3,3,3,2,2,1};
int k = 2;
// get first element
int element=array[0];
// set tot to 1
int tot=1;
for (int i = 1; i < array.length; i++) {
// if values are same , increment tot
if (element==array[i]) {
tot++;
}else {
// set element to new found value and tot = 1
element=array[i];
tot=1;
}
// when any elements exceeds the k limit
// print element and stop the loop
if (tot>k) {
System.out.println(array[i]);
break;
}
}
// to print -1 if tot > k never reached
if (!(tot>k)) {
System.out.println(-1);
}

adding 1 to an integer within println statement in Java [duplicate]

Is there a difference in ++i and i++ in a for loop? Is it simply a syntax thing?

a++ is known as postfix.
add 1 to a, returns the old value.
++a is known as prefix.
add 1 to a, returns the new value.
C#:
string[] items = {"a","b","c","d"};
int i = 0;
foreach (string item in items)
{
Console.WriteLine(++i);
}
Console.WriteLine("");
i = 0;
foreach (string item in items)
{
Console.WriteLine(i++);
}
Output:
1
2
3
4
0
1
2
3
foreach and while loops depend on which increment type you use. With for loops like below it makes no difference as you're not using the return value of i:
for (int i = 0; i < 5; i++) { Console.Write(i);}
Console.WriteLine("");
for (int i = 0; i < 5; ++i) { Console.Write(i); }
0 1 2 3 4
0 1 2 3 4
If the value as evaluated is used then the type of increment becomes significant:
int n = 0;
for (int i = 0; n < 5; n = i++) { }

Pre-increment ++i increments the value of i and evaluates to the new incremented value.
int i = 3;
int preIncrementResult = ++i;
Assert( preIncrementResult == 4 );
Assert( i == 4 );
Post-increment i++ increments the value of i and evaluates to the original non-incremented value.
int i = 3;
int postIncrementResult = i++;
Assert( postIncrementtResult == 3 );
Assert( i == 4 );
In C++, the pre-increment is usually preferred where you can use either.
This is because if you use post-increment, it can require the compiler to have to generate code that creates an extra temporary variable. This is because both the previous and new values of the variable being incremented need to be held somewhere because they may be needed elsewhere in the expression being evaluated.
So, in C++ at least, there can be a performance difference which guides your choice of which to use.
This is mainly only a problem when the variable being incremented is a user defined type with an overridden ++ operator. For primitive types (int, etc) there's no performance difference. But, it's worth sticking to the pre-increment operator as a guideline unless the post-increment operator is definitely what's required.
There's some more discussion here.
In C++ if you're using STL, then you may be using for loops with iterators. These mainly have overridden ++ operators, so sticking to pre-increment is a good idea. Compilers get smarter all the time though, and newer ones may be able to perform optimizations that mean there's no performance difference - especially if the type being incremented is defined inline in header file (as STL implementations often are) so that the compiler can see how the method is implemented and can then know what optimizations are safe to perform. Even so, it's probably still worth sticking to pre-increment because loops get executed lots of times and this means a small performance penalty could soon get amplified.
In other languages such as C# where the ++ operator can't be overloaded there is no performance difference. Used in a loop to advance the loop variable, the pre and post increment operators are equivalent.
Correction: overloading ++ in C# is allowed. It seems though, that compared to C++, in C# you cannot overload the pre and post versions independently. So, I would assume that if the result of calling ++ in C# is not assigned to a variable or used as part of a complex expression, then the compiler would reduce the pre and post versions of ++ down to code that performs equivalently.

In C# there is no difference when used in a for loop.
for (int i = 0; i < 10; i++) { Console.WriteLine(i); }
outputs the same thing as
for (int i = 0; i < 10; ++i) { Console.WriteLine(i); }
As others have pointed out, when used in general i++ and ++i have a subtle yet significant difference:
int i = 0;
Console.WriteLine(i++); // Prints 0
int j = 0;
Console.WriteLine(++j); // Prints 1
i++ reads the value of i then increments it.
++i increments the value of i then reads it.

The question is:
Is there a difference in ++i and i++ in a for loop?
The answer is: No.
Why does each and every other answer have to go into detailed explanations about pre and post incrementing when this is not even asked?
This for-loop:
for (int i = 0; // Initialization
i < 5; // Condition
i++) // Increment
{
Output(i);
}
Would translate to this code without using loops:
int i = 0; // Initialization
loopStart:
if (i < 5) // Condition
{
Output(i);
i++ or ++i; // Increment
goto loopStart;
}
Now does it matter if you put i++ or ++i as increment here? No it does not as the return value of the increment operation is insignificant. i will be incremented AFTER the code's execution that is inside the for loop body.

Since you ask about the difference in a loop, i guess you mean
for(int i=0; i<10; i++)
...;
In that case, you have no difference in most languages: The loop behaves the same regardless of whether you write i++ and ++i. In C++, you can write your own versions of the ++ operators, and you can define separate meanings for them, if the i is of a user defined type (your own class, for example).
The reason why it doesn't matter above is because you don't use the value of i++. Another thing is when you do
for(int i=0, a = 0; i<10; a = i++)
...;
Now, there is a difference, because as others point out, i++ means increment, but evaluate to the previous value, but ++i means increment, but evaluate to i (thus it would evaluate to the new value). In the above case, a is assigned the previous value of i, while i is incremented.

One (++i) is preincrement, one (i++) is postincrement. The difference is in what value is immediately returned from the expression.
// Psuedocode
int i = 0;
print i++; // Prints 0
print i; // Prints 1
int j = 0;
print ++j; // Prints 1
print j; // Prints 1
Edit: Woops, entirely ignored the loop side of things. There's no actual difference in for loops when it's the 'step' portion (for(...; ...; )), but it can come into play in other cases.

As this code shows (see the dissambled MSIL in the comments), the C# 3 compiler makes no distinction between i++ and ++i in a for loop. If the value of i++ or ++i were being taken, there would definitely be a difference (this was compiled in Visutal Studio 2008 / Release Build):
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace PreOrPostIncrement
{
class Program
{
static int SomethingToIncrement;
static void Main(string[] args)
{
PreIncrement(1000);
PostIncrement(1000);
Console.WriteLine("SomethingToIncrement={0}", SomethingToIncrement);
}
static void PreIncrement(int count)
{
/*
.method private hidebysig static void PreIncrement(int32 count) cil managed
{
// Code size 25 (0x19)
.maxstack 2
.locals init ([0] int32 i)
IL_0000: ldc.i4.0
IL_0001: stloc.0
IL_0002: br.s IL_0014
IL_0004: ldsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0009: ldc.i4.1
IL_000a: add
IL_000b: stsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0010: ldloc.0
IL_0011: ldc.i4.1
IL_0012: add
IL_0013: stloc.0
IL_0014: ldloc.0
IL_0015: ldarg.0
IL_0016: blt.s IL_0004
IL_0018: ret
} // end of method Program::PreIncrement
*/
for (int i = 0; i < count; ++i)
{
++SomethingToIncrement;
}
}
static void PostIncrement(int count)
{
/*
.method private hidebysig static void PostIncrement(int32 count) cil managed
{
// Code size 25 (0x19)
.maxstack 2
.locals init ([0] int32 i)
IL_0000: ldc.i4.0
IL_0001: stloc.0
IL_0002: br.s IL_0014
IL_0004: ldsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0009: ldc.i4.1
IL_000a: add
IL_000b: stsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0010: ldloc.0
IL_0011: ldc.i4.1
IL_0012: add
IL_0013: stloc.0
IL_0014: ldloc.0
IL_0015: ldarg.0
IL_0016: blt.s IL_0004
IL_0018: ret
} // end of method Program::PostIncrement
*/
for (int i = 0; i < count; i++)
{
SomethingToIncrement++;
}
}
}
}

Here is a Java-Sample and the Byte-Code, post- and preIncrement show no difference in Bytecode:
public class PreOrPostIncrement {
static int somethingToIncrement = 0;
public static void main(String[] args) {
final int rounds = 1000;
postIncrement(rounds);
preIncrement(rounds);
}
private static void postIncrement(final int rounds) {
for (int i = 0; i < rounds; i++) {
somethingToIncrement++;
}
}
private static void preIncrement(final int rounds) {
for (int i = 0; i < rounds; ++i) {
++somethingToIncrement;
}
}
}
And now for the byte-code (javap -private -c PreOrPostIncrement):
public class PreOrPostIncrement extends java.lang.Object{
static int somethingToIncrement;
static {};
Code:
0: iconst_0
1: putstatic #10; //Field somethingToIncrement:I
4: return
public PreOrPostIncrement();
Code:
0: aload_0
1: invokespecial #15; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]);
Code:
0: sipush 1000
3: istore_1
4: sipush 1000
7: invokestatic #21; //Method postIncrement:(I)V
10: sipush 1000
13: invokestatic #25; //Method preIncrement:(I)V
16: return
private static void postIncrement(int);
Code:
0: iconst_0
1: istore_1
2: goto 16
5: getstatic #10; //Field somethingToIncrement:I
8: iconst_1
9: iadd
10: putstatic #10; //Field somethingToIncrement:I
13: iinc 1, 1
16: iload_1
17: iload_0
18: if_icmplt 5
21: return
private static void preIncrement(int);
Code:
0: iconst_0
1: istore_1
2: goto 16
5: getstatic #10; //Field somethingToIncrement:I
8: iconst_1
9: iadd
10: putstatic #10; //Field somethingToIncrement:I
13: iinc 1, 1
16: iload_1
17: iload_0
18: if_icmplt 5
21: return
}

There is no difference if you are not using the value after increment in the loop.
for (int i = 0; i < 4; ++i){
cout<<i;
}
for (int i = 0; i < 4; i++){
cout<<i;
}
Both the loops will print 0123.
But the difference comes when you uses the value after increment/decrement in your loop as below:
Pre Increment Loop:
for (int i = 0,k=0; i < 4; k=++i){
cout<<i<<" ";
cout<<k<<" ";
}
Output:
0 0
1 1
2 2
3 3
Post Increment Loop:
for (int i = 0, k=0; i < 4; k=i++){
cout<<i<<" ";
cout<<k<<" ";
}
Output:
0 0
1 0
2 1
3 2
I hope the difference is clear by comparing the output. Point to note here is the increment/decrement is always performed at the end of the for loop and hence the results can be explained.

Yes, there is. The difference is in the return value. The return value of "++i" will be the value after incrementing i. The return of "i++" will be the value before incrementing. This means that code that looks like the following:
int a = 0;
int b = ++a; // a is incremented and the result after incrementing is saved to b.
int c = a++; // a is incremented again and the result before incremening is saved to c.
Therefore, a would be 2, and b and c would each be 1.
I could rewrite the code like this:
int a = 0;
// ++a;
a = a + 1; // incrementing first.
b = a; // setting second.
// a++;
c = a; // setting first.
a = a + 1; // incrementing second.

There is no actual difference in both cases 'i' will be incremented by 1.
But there is a difference when you use it in an expression, for example:
int i = 1;
int a = ++i;
// i is incremented by one and then assigned to a.
// Both i and a are now 2.
int b = i++;
// i is assigned to b and then incremented by one.
// b is now 2, and i is now 3

There is more to ++i and i++ than loops and performance differences. ++i returns a l-value and i++ returns an r-value. Based on this, there are many things you can do to ( ++i ) but not to ( i++ ).
1- It is illegal to take the address of post increment result. Compiler won't even allow you.
2- Only constant references to post increment can exist, i.e., of the form const T&.
3- You cannot apply another post increment or decrement to the result of i++, i.e., there is no such thing as I++++. This would be parsed as ( i ++ ) ++ which is illegal.
4- When overloading pre-/post-increment and decrement operators, programmers are encouraged to define post- increment/decrement operators like:
T& operator ++ ( )
{
// logical increment
return *this;
}
const T operator ++ ( int )
{
T temp( *this );
++*this;
return temp;
}

It boggles my mind why so may people write the increment expression in for-loop as i++.
In a for-loop, when the 3rd component is a simple increment statement, as in
for (i=0; i<x; i++)
or
for (i=0; i<x; ++i)
there is no difference in the resulting executions.

As #Jon B says, there is no difference in a for loop.
But in a while or do...while loop, you could find some differences if you are making a comparison with the ++i or i++
while(i++ < 10) { ... } //compare then increment
while(++i < 10) { ... } //increment then compare

In javascript due to the following i++ may be better to use:
var i=1;
alert(i++); // before, 1. current, 1. after, 2.
alert(i); // before, 2. current, 2. after, 2.
alert(++i); // before, 2. current, 3 after, 3.
While arrays (I think all) and some other functions and calls use 0 as a starting point you would have to set i to -1 to make the loop work with the array when using ++i.
When using i++ the following value will use the increased value. You could say i++ is the way humans count, cause you can start with a 0.

To understand what a FOR loop does
The image above shows that FOR can be converted to WHILE, as they eventually have totally the same assembly code (at least in gcc). So we can break down FOR into a couple of pieces, to undertand what it does.
for (i = 0; i < 5; ++i) {
DoSomethingA();
DoSomethingB();
}
is equal to the WHILE version
i = 0; //first argument (a statement) of for
while (i < 5 /*second argument (a condition) of for*/) {
DoSomethingA();
DoSomethingB();
++i; //third argument (another statement) of for
}
It means that you can use FOR as a simple version of WHILE:
The first argument of FOR (int i) is executed, outside, before the loop.
The third argument of FOR (i++ or ++i) is executed, inside, in the last line of the loop.
TL:DR: no matter whether i++ or ++i, we know that when they are standalone, they make no difference but +1 on themselves.
In school, they usually teach the i++ way, but there are also lots of people prefer the ++i way due to several reasons.
NOTE: In the past, i++ has very little impact on the performance, as it does not only plus one by itself, but also keeps the original value in the register. But for now, it makes no difference as the compiler makes the plus one part the same.

There can be a difference for loops. This is the practical application of post/pre-increment.
int i = 0;
while(i++ <= 10) {
Console.Write(i);
}
Console.Write(System.Environment.NewLine);
i = 0;
while(++i <= 10) {
Console.Write(i);
}
Console.ReadLine();
While the first one counts to 11 and loops 11 times, the second does not.
Mostly this is rather used in a simple while(x-- > 0 ) ; - - Loop to iterate for example all elements of an array (exempting foreach-constructs here).

Yes, there is a difference between ++i and i++ in a for loop, though in unusual use cases; when a loop variable with increment/decrement operator is used in the for block or within the loop test expression, or with one of the loop variables. No it is not simply a syntax thing.
As i in a code means evaluate the expression i and the operator does not mean an evaluation but just an operation;
++i means increment value of i by 1 and later evaluate i,
i++ means evaluate i and later increment value of i by 1.
So, what are obtained from each two expressions differ because what is evaluated differs in each. All same for --i and i--
For example;
let i = 0
i++ // evaluates to value of i, means evaluates to 0, later increments i by 1, i is now 1
0
i
1
++i // increments i by 1, i is now 2, later evaluates to value of i, means evaluates to 2
2
i
2
In unusual use cases, however next example sounds useful or not does not matter, it shows a difference
for(i=0, j=i; i<10; j=++i){
console.log(j, i)
}
for(i=0, j=i; i<10; j=i++){
console.log(j, i)
}

In certain situations ++i and i+1 might give different results, same goes for --i, i - 1 etc.
This is not because there is a flaw in how increment and decrement operators work but because of a little fact sometimes overlooked by new programmers.
As a rule of thumb do not use inc/dec inside array's square brackets. For example, I won't do something like arr[++i] in place of arr[i + 1]. Though both would get us the same value of i, there is something that we overlooked here.
If a loop condition is based on i's value for execution then replacing arr[i + 1] with arr[++i] would result in error. Why?
Let say i = 5, then arr[i + 1] would mean arr[6] and arr[++i] although would mean arr[6] but would also change the value of i to 6 and this might not be something that we want to do. We might not want to change the value of i but due to a simple ++/-- operator, we changed the value.
So be careful when using ++/-- operators.
I hope, I was able to make my point easy for understanding.

For i's of user-defined types, these operators could (but should not) have meaningfully different sematics in the context of a loop index, and this could (but should not) affect the behavior of the loop described.
Also, in c++ it is generally safest to use the pre-increment form (++i) because it is more easily optimized. (Scott Langham beat me to this tidbit. Curse you, Scott)

I dont know for the other languages but in Java ++i is a prefix increment which means: increase i by 1 and then use the new value of i in the expression in which i resides, and i++ is a postfix increment which means the following: use the current value of i in the expression and then increase it by 1.
Example:
public static void main(String [] args){
int a = 3;
int b = 5;
System.out.println(++a);
System.out.println(b++);
System.out.println(b);
}
and the output is:
4
5
6

i++ ; ++i ; both are similar as they are not used in an expression.
class A {
public static void main (String []args) {
int j = 0 ;
int k = 0 ;
++j;
k++;
System.out.println(k+" "+j);
}}
prints out : 1 1

Java Incremental operator query (++i and i++) [duplicate]

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I have the following code:
public class Book {
private static int sample1(int i) {
return i++;
}
private static int sample2(int j) {
return ++j;
}
public static void main(String[] arguments){
int i = 0;
int j = 0;
System.out.println(sample1(i++)); //0
System.out.println(sample1(++i)); //1
System.out.println(sample2(j++));//1
System.out.println(sample2(++j));//2
System.out.println(i);//2
System.out.println(j);//2
}
}
My expected output is in comments. The actual output is below:
0
2
1
3
2
2
I'm getting confused with the function calls and incemental operator. Can someone kindly explain the actual result?

Since sample1 and sample2 are just modifying their own local variables i and j (not those of the calling method), it's clearer if we rewrite them without those modifications:
private static int sample1(int i) {
return i; // was 'i++', which evaluates to the old i
}
private static int sample2(int j) {
return j + 1; // was '++j', which evaluates to j after incrementing
}
At which point it's straightforward to just substitute them in place — sample1(...) becomes ..., and sample2(...) becomes ... + 1:
int i = 0;
int j = 0;
System.out.println(i++);
System.out.println(++i);
System.out.println((j++) + 1);
System.out.println((++j) + 1);
System.out.println(i);
System.out.println(j);
We can make this a bit clearer by separating the incrementations into their own commands. i++ evaluates to the original value of i, so it's like incrementing i after running the surrounding command; ++i, by contrast, is like incrementing i before running the surrounding command. So we get:
int i = 0;
int j = 0;
System.out.println(i);
i++;
++i;
System.out.println(i);
System.out.println(j + 1);
j++;
++j;
System.out.println(j + 1);
System.out.println(i);
System.out.println(j);
. . . at which point it should be straightforward to trace through and see what it will output.
Does that all make sense?

First of all you need to know the difference between x++ and ++X;
In case of x++ :
First the current value will be used and it will be incremented next.
That means you will get the present value of x for the operation and if you
use x next time will get the incremented value;
In case of ++x :
First the current value will be incremented and it will be used (the incremented value) next, that means you will get the incremented value
at this operation and for other after this operation.
Now lets split the code and discuss them separately
method: sample1() :
private static int sample1(int i) {
return i++;
}
This method will take a int and return it first and then try to increment but as after returning the variable i will go out of scope so it will never be
incremented at all. exp in: 10-> out 10
method: sample2() :
private static int sample2(int j) {
return ++j;
}
This method will take a int and increment it first and then return it. exp in: 10-> out 11
In both case only the variables will change locally, that means if you call from main method the variables of main method will remain unaffected by the change
(as the sample1() and sample2() are making copy of the variables)
Now for the code of the main method
System.out.println(sample1(i++)); // it's giving sample1() `i=0` then making `i=1`
// so sample1() will return 0 too;
System.out.println(sample1(++i)); // it's making `i=2` and then giving sample1() `i=2`
// so sample1() will return 2;
System.out.println(sample2(j++)); // it's giving sample2() `j=0` then making `j=1`
// so sample2() will return 1;
System.out.println(sample2(++j)); // it's making `j=2` giving sample2() `j=2` then
// so sample2() will return 3;

You're experiencing the fun of prefix and postfix operators.
The prefix operator, ++i increments the variable i by one before using it in the expression, where the postfix operator (i++) uses i in the expression before incrementing it.
This means that your method sample1 doesn't do anything; it evaluates the expression containing i, but because that expression is a return statement, the local variable i goes out of scope, and we can't modify it anymore.
sample2, by contrast, increments the local copy of j before returning it, which is why you print higher values of j than you expect.

Easy:
1) First call:
a) Provide i (==0) to sample1(), which returns 0 (then increments the argument i, and that gets discarded).
b) increments i because of i++. i is now 1.
c) Prints the function result: 0.
2) Second call:
a) Increments i because of ++i. i is now 2.
b) Provide i (==2) to sample1(), which returns 2 (then increments the argument i, and that gets discarded)
c) Prints the function result: 2.
3) Third call:
a) Provide j (==0) to sample2(), which increments the argument, and therefore returns 1.
b) increments j because of j++. j is now 1.
c) Prints the function result: 1.
4) Fourth call:
a) Increments j because of ++j. j is now 2.
b) Provide j (==2) to sample2(), which increments the argument, and therefore returns 3.
c) Prints the function result: 3.
5 & 6) Fifth and Sixth call:
a) Prints the value of j: 2.
The key to remember here is that i++ increments the variable after passing it as an argument, whereas ++i increments the variable before passing it as an argument.
Hope this helps

1st print
before call: i = 0
increment after call
sample1 is called with a value of 0
sample 1 returns 0, the increment is discarded
after call: i = 1
2nd print
before call: i = 1
increment before call
sample1 is called with a value of 2
sample1 returns 2, the increment is discarded
after call: i = 2
3rd print
before call: j = 0
increment after call
sample2 is called with a value of 0
sample2 the increments 0 to 1, returns it
1 is printed
increment j to 1
after call: j = 1
4th print
before call: j = 1
increment before call
increment j to 2
sample2 is called with a value of 2
sample2 the increments 2 to 3, returns it
3 is printed
after call: j = 2
5th print
prints i
2 is printed
6th print
prints j
2 is printed

Both of them increase the variable i by one like i = i + 1;
The difference is that :
++i increments the value first and then return it
i++ return the value first and then increments it
This behavior difference doesn’t matter in a for loop.
If you want to know the difference try this :
int x = 0;
int y = x++;
int x = 0;
int y = ++x;
Here x++ returns the value then increments it but ++x first increments the value then returns that value

Form your example,
private static int sample1(int i) {
return i++;
}
private static int sample2(int j) {
return ++j;
}
public static void main(String[] arguments)
{
int i = 0;
int j = 0;
System.out.println(sample1(i++)); //0
System.out.println(sample1(++i)); //1
System.out.println(sample2(j++));//1
System.out.println(sample2(++j));//2
System.out.println(j);//2
System.out.println(j);//2
}
i = 0; sample1(i++) -> it pass '0'->in sample1 return i++ so, 0(++)
Here it returns 0 but incremented to 1 , so println is = 0 but finally i takes 1
i = 1; sample1(++i) -> it pass '2'-> in sample1 return i++ so, 2(++)
Here it returns 2, so println is = 2
j = 0; sample2(j++) -> it pass '0'-> in sample2 return ++j so, (++)0
Here it returns 1, so println is = 1.
j = 1; sample2(++j) -> it pass ++1 => 2, in sample2 return ++j so, (++)2
Here it returns 3, so println is = 3. But increment is ended with in sample2, not in
main so j still holds 2.
j = 2
j = 2

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