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How to add numbers without +

I was asked this question on the interview. I didn't answer and actually I don't understand how it works.
int add(int x, int y)
{
while (y != 0)
{
int carry = x & y;
x = x ^ y;
y = carry << 1;
}
return x;
}
I'm not asking why does it produce a correct answer... First of all, why does the algorithm eventually stop? To me it's not that obvious.
In order for it to stop, carry has to become 0. Can't someone explain it in a nutshell?

line 1 : int carry = x & y;
line 2 : x = x ^ y;
line 3 : y = carry << 1;
if x = 1; y = 2;
Binary for each number:
0 = 00
1 = 01
2 = 10
3 = 11
for line 1 code,
& (bitwise AND)
Binary AND Operator copies a bit to the result if it exists in both operands
x is 1 => 01
y is 2 => 10
result carry is => 00 (0)
for line 2 code,
^ (bitwise XOR)
Binary XOR Operator copies the bit if it is set in one operand but not both.
x is 1 => 01
y is 2 => 10
result x is => 11 (3)
for line 3 code,
variable carry needs to shift left for 1 bit,
so now carry is 0 => 00 and shift 1 bit left means carry is now 0. The result y is (0). And while loop stop because y is 0 now.
The final result for x is 3.
Hope this will help you.

Let's take an example:
x=13(1101)
y=9(1001)
Loop 1:
-----------------
y!=0 -> carry=(1101)&(1001)=1001(9) [AND Op]
x=(1101)^(1001)=0100(4) [XOR Op]
y=carry<<1 -> y=(carry)x2=10010(18)
Loop 2:
-----------------
y!=0 -> carry=(0100)&(10010)=00000(0)
x=(0100)^(10010)=10110(22)
y=carry<<1 -> y=0
loop terminated.
therefore,x is 22.So,x^y store the sum part and x&y store the carry part,and then carry(x&y) is shifted to match the digit with x^y,and finnally XOR them and store into x.
x is the resultant.

In a nutshell its about using y (and the "carries/x&y" it becomes) to modify x until it becomes the sum of both ints. For example,
y=1 (....0001), x=anything (either .....0 or .....1)
if x ends with 0, x&y=0
//x^y = x becomes ....001 (thereby adding 1)
//since x&y=0 the loop stops
if x ends with 1, x&y=1
//x^y = x
//since y= x&y<<1, new y=(.....000010)
if x ends with 01, x&y=0
//x^y = x becomes ....010 (thereby adding 1)
//since x&y=0 the loop stops
if x ends with 11, x&y=1
//x^y = .....01
//since y= x&y<<1, new y=(......000100)
if x ends with 011
//stuff happens and x becomes ....100 (thereby adding 1)
//loop stops
if x ends with 111
//...
//if x ends with 111111, x becomes ....1000000 (thereby adding 1)
//if x ends with 1111111, x becomes ....10000000 (thereby adding 1)
//if x ends with 11111111, x becomes ....100000000 (thereby adding 1)
//well you get the idea
The same logic is applicable to all values of y, and is not that much different from normal addition only that there are now 2 possible digits (0 and 1) instead of the usual 10 (0 to 9).

Multiple assignment on one line not working as expected

I'm trying to swap two ints - x and y in the example, and do it in one line without a library function.
So I started with this:
int x = 4;
int y = 3;
System.out.println(x);
System.out.println(y);
x ^= y;
System.out.println(x);
System.out.println(y);
y ^= x;
System.out.println(x);
System.out.println(y);
x ^= y;
System.out.println(x);
System.out.println(y);
The output was 4, 3, 7, 3, 7, 4, 3, 4 as expected. All good so far.
Next up was this:
int x = 4;
int y = 3;
System.out.println(x);
System.out.println(y);
y ^= (x ^= y);
System.out.println(x);
System.out.println(y);
x ^= y;
System.out.println(x);
System.out.println(y);
The output was 4, 3, 7, 4, 3, 4 as expected once again. Still good so far.
Then finally this:
int x = 4;
int y = 3;
System.out.println(x);
System.out.println(y);
x ^= (y ^= (x ^= y));
System.out.println(x);
System.out.println(y);
At this stage the output became 4, 3, 0, 4. Now I know that the 0 is a result of 4 ^ 4 because the x assignment wasn't complete at that time - why is this happening? Why doesn't the x ^= y actually assign 7 to the x variable so that it becomes 7 ^ 4 for the last assignment?

Let's try to expand your last expression.
It evaluates to,
x = x^(y = y^ (x = x^y));
Note that expressions are evaluated from left to right,
it becomes,
x = 4 ^ (y = 3 ^ (x = 4 ^ 3));
Now, the problem has become obvious. Right?
Edit:
To clear come confusion, let me try to explain what I mean by evaluating from left to right.
int i = 1;
s = i + (i = 2) + i;
Now, the expression will evaluate to,
s = 1 + 2 + 2;
Notice that i on the left of assignment was 1, but on the right of assignment (and on the assigment) was evaluated to 2, because the evaluation is from left to right, when it came to the 2nd and 3rd part of the expression, is value was 2.

The order of evaluation is defined in chapter 15 of the JLS. Item 15.7.1 says:
If the operator is a compound-assignment operator (§15.26.2), then evaluation of the left-hand operand includes both remembering the variable that the left-hand operand denotes and fetching and saving that variable's value for use in the implied binary operation.
To explain further, they have two examples of computations that involve assignments. Here the assignment is on the left hand of the operator:
int i = 2;
int j = (i=3) * i;
System.out.println(j);
And they specifically say that the result is 9 and is not allowed to be 6. That is, the (i=3) is both calculated as 3 and i is assigned 3 before being multiplied with itself.
But in the second example:
int a = 9;
a += (a = 3); // first example
System.out.println(a);
int b = 9;
b = b + (b = 3); // second example
System.out.println(b);
The JLS specifies that both prints should produce 12, and are not allowed to produce 6. That is, because the assignment to b is on the right side, the value of the left b (or the implicit left a in the += operation), the value before that assignment is fetched and saved first, and only then the operation inside the parentheses is performed.
Internally, expressions are broken down into JVM operations that push and pop values onto an "operand stack". If you think about it like that - that in b = b + (b=3) The value of b is first pushed onto the operand stack, then the (b=3) is performed and its value is then added to the value popped from the stack (old value of b), it will make sense. At this point, the left hand b just stands for "What the value of b was when it was pushed on the stack" and not for the "current value of b".

Lets divide and compute.
The innermost parenthesis executes first and after all the parenthesis resolved, then the expression executes from left to right.
x ^= (y ^= (x ^= y)); // initial statement
x = x^(y = y^ (x = x^y)); //equals to
() have the highest precedence
x = x^(y = y^ (x = 3^4)); // first highest precedence ()
x = x^(y = y ^ (x = 7)); // still the first x is 3
x = 4 ^(y = 3 ^ (x = 7)); // now 3 ^ 7 =4
x = 4 ^ 4; // now 3 ^ 7 =4
x= 0;

Java Increment / Decrement Operators - How they behave, what's the functionality?

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.

Java executes arithmetic Expression wrong?

I don`t understand how Java is progressing this arithmetic expression
int x = 1;
int y = 1;
x += y += x += y;
System.out.println("x=" + x + " y=" + y);
With Java I get x = 4 and y = 3. But in C, Perl, Php I get x=5 and y = 3
On the paper I also get x = 5 and y = 3

This is the nasty part, obviously:
x += y += x += y;
This is executed as:
int originalX = x; // Used later
x = x + y; // Right-most x += y
y = y + x; // Result of "x += y" is the value stored in x
x = originalX + y; // Result of "y += x" is the value stored in y
So:
x y
(Start) 1 1
x = x + y 2 1
y = y + x 2 3
x = originalX + y 4 3
The important part is the use of originalX here. The compound assignment is treated as:
x = x + y
and the first operand of + is evaluated before the second operand... which is why it takes the original value of x, not the "latest" one.
From JLS section 15.16.2:
If the left-hand operand expression is not an array access expression, then:
First, the left-hand operand is evaluated to produce a variable. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason; the right-hand operand is not evaluated and no assignment occurs.
Otherwise, the value of the left-hand operand is saved and then the right-hand operand is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason and no assignment occurs.
When in doubt, consult the language specification - and never assume that just because two languages behave differently, one of them is "wrong". So long as the actual behaviour matches the specified behaviour for each language, all is well - but it does mean you need to understand the behaviour of each language you work with, of course.
That said, you should clearly avoid horrible code like this in the first place.

it starts from right to left
1. x += y gives x = 2, y = 1
2. y += ( x +=y ) gives x = 2, y = 3
3. x += ( y += x += y ) gives x = 4, y = 3
so, it's all consistent

How does the following statement work?

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