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Why does this go into an infinite loop?

I have the following code:

public class Tests {
    public static void main(String[] args) throws Exception {
        int x = 0;
        while(x<3) {
            x = x++;
            System.out.println(x);
        }
    }
}

We know he should have writen just x++ or x=x+1, but on x = x++ it should first attribute x to itself, and later increment it. Why does x continue with 0 as value?

--update

Here's the bytecode:

public class Tests extends java.lang.Object{
public Tests();
  Code:
   0:   aload_0
   1:   invokespecial   #1; //Method java/lang/Object."<init>":()V
   4:   return

public static void main(java.lang.String[])   throws java.lang.Exception;
  Code:
   0:   iconst_0
   1:   istore_1
   2:   iload_1
   3:   iconst_3
   4:   if_icmpge   22
   7:   iload_1
   8:   iinc    1, 1
   11:  istore_1
   12:  getstatic   #2; //Field java/lang/System.out:Ljava/io/PrintStream;
   15:  iload_1
   16:  invokevirtual   #3; //Method java/io/PrintStream.println:(I)V
   19:  goto    2
   22:  return

}

I'll read about the instructions to try to understand...

I suspect what's happening is: 1. load x into a register (=0); 2. increment x (x=1); 3. save register value into x (x=0). In C/C++ this would be undefined behaviour because there's no formal sequence point to define the order of 2 and 3. Hopefully someone can quote you something equivalent from the Java spec.
We tried this in C++ to see what would happen, and it prints 1,2,3 and exits. I did not expect that. I assume it is compiler dependent, since it is undefined behavior. We used the gnu g++.
@saj x++ is post-increment; x= is assignment of result; the result of x++ is the original x (and there is a side-effect of increment, but that doesn't change the result), So this can be interpreted as var tmp = x; x++; x = tmp;
Now that I have a popular question I'm regretful, the (repeated) answers don't stop coming, even after the correct answer was chosen. My "Recent Activity" screen is full of the same answers, and coming more...
@Rob Vermeulen you may want to read the full question before make a comment.. ;) This was a code made by a student of mine, and I was curious why of this behavior.

D
Dan Tao

Note: Originally I posted C# code in this answer for purposes of illustration, since C# allows you to pass int parameters by reference with the ref keyword. I've decided to update it with actual legal Java code using the first MutableInt class I found on Google to sort of approximate what ref does in C#. I can't really tell if that helps or hurts the answer. I will say that I personally haven't done all that much Java development; so for all I know there could be much more idiomatic ways to illustrate this point.

Perhaps if we write out a method to do the equivalent of what x++ does it will make this clearer.

public MutableInt postIncrement(MutableInt x) {
    int valueBeforeIncrement = x.intValue();
    x.add(1);
    return new MutableInt(valueBeforeIncrement);
}

Right? Increment the value passed and return the original value: that's the definition of the postincrement operator.

Now, let's see how this behavior plays out in your example code:

MutableInt x = new MutableInt();
x = postIncrement(x);

postIncrement(x) does what? Increments x, yes. And then returns what x was before the increment. This return value then gets assigned to x.

So the order of values assigned to x is 0, then 1, then 0.

This might be clearer still if we re-write the above:

MutableInt x = new MutableInt();    // x is 0.
MutableInt temp = postIncrement(x); // Now x is 1, and temp is 0.
x = temp;                           // Now x is 0 again.

Your fixation on the fact that when you replace x on the left side of the above assignment with y, "you can see that it first increments x, and later attributes it to y" strikes me as confused. It is not x that is being assigned to y; it is the value formerly assigned to x. Really, injecting y makes things no different from the scenario above; we've simply got:

MutableInt x = new MutableInt();    // x is 0.
MutableInt y = new MutableInt();    // y is 0.
MutableInt temp = postIncrement(x); // Now x is 1, and temp is 0.
y = temp;                           // y is still 0.

So it's clear: x = x++ effectively does not change the value of x. It always causes x to have the values x0, then x0 + 1, and then x0 again.

Update: Incidentally, lest you doubt that x ever gets assigned to 1 "between" the increment operation and the assignment in the example above, I've thrown together a quick demo to illustrate that this intermediate value does indeed "exist," though it will never be "seen" on the executing thread.

The demo calls x = x++; in a loop while a separate thread continuously prints the value of x to the console.

public class Main {
    public static volatile int x = 0;

    public static void main(String[] args) {
        LoopingThread t = new LoopingThread();
        System.out.println("Starting background thread...");
        t.start();

        while (true) {
            x = x++;
        }
    }
}

class LoopingThread extends Thread {
    public @Override void run() {
        while (true) {
            System.out.println(Main.x);
        }
    }
}

Below is an excerpt of the above program's output. Notice the irregular occurrence of both 1s and 0s.

Starting background thread...
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
0
1

You don't need to create a class to pass by reference in java (though that would certainly work). You can use the Integer class, which is part of the standard library, and it even has the benefit of being auto-boxed to and from int almost transparently.
@rmeador Integer is immutable, so you still couldn't change its value. AtomicInteger, however, is mutable.
@Dan: By the way, x in your last example must be declared volatile, otherwise it's an undefined behaviour and seeing 1s is implementation specific.
@burkestar: I don't think that link is quite appropriate in this case, since it's a Java question and (unless I'm mistaken) the behavior is actually undefined in C++.
@Tom Brito - in C it's not defined... the ++ could be done before or after assignment. Practically speaking, there might a compiler that does the same thing as Java, but you wouldn't want to bet on it.
a
axtavt

x = x++ works in the following way:

First it evaluates expression x++. Evaluation of this expression produces an expression value (which is the value of x before increment) and increments x.

Later it assigns the expression value to x, overwriting incremented value.

So, the sequence of events looks like follows (it's an actual decompiled bytecode, as produced by javap -c, with my comments):

8:   iload_1         // Remember current value of x in the stack
   9:   iinc    1, 1    // Increment x (doesn't change the stack)
   12:  istore_1        // Write remebered value from the stack to x

For comparison, x = ++x:

8:   iinc    1, 1    // Increment x
   11:  iload_1         // Push value of x onto stack
   12:  istore_1        // Pop value from the stack to x

if you make a test, you can see that it first increments, and later attributes. So it should not attribute zero.
@Tom that's the point, though - because this is all a single sequence it's doing things in a non-obvious (and probably undefined) order. By attempting to test this you're adding a sequence point and getting different behaviour.
@Rep It may not be defined in C or C++, but in Java, it is well defined.
@Jaydee - almost... the purpose of the standard is that standard conforming code will operate the same way :) At any rate, there was (and maybe still is) an advantage to not specifying sequence points under every possible circumstance in C, but it's not really an advantage in Java.
Y
Yoon5oo

This happens because the value of x doesn't get incremented at all.

x = x++;

is equivalent to

int temp = x;
x++;
x = temp;

Explanation:

Let's look at the byte code for this operation. Consider a sample class:

class test {
    public static void main(String[] args) {
        int i=0;
        i=i++;
    }
}

Now running the class disassembler on this we get:

$ javap -c test
Compiled from "test.java"
class test extends java.lang.Object{
test();
  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_0
   1:    istore_1
   2:    iload_1
   3:    iinc    1, 1
   6:    istore_1
   7:    return
}

Now the Java VM is stack based which means for each operation, the data will be pushed onto the stack and from the stack, the data will pop out to perform the operation. There is also another data structure, typically an array to store the local variables. The local variables are given ids which are just the indexes to the array.

Let us look at the mnemonics in main() method:

iconst_0: The constant value 0 is pushed on to the stack.

istore_1: The top element of the stack is popped out and stored in the local variable with index 1 which is x.

iload_1 : The value at the location 1 that is the value of x which is 0, is pushed into the stack.

iinc 1, 1 : The value at the memory location 1 is incremented by 1. So x now becomes 1.

istore_1 : The value at the top of the stack is stored to the memory location1. That is 0 is assigned to x overwriting its incremented value.

Hence the value of x does not change resulting in the infinite loop.


Actually it gets incremented (thats the meaning of ++), but the variable gets overwritten later.
int temp = x; x = x + 1; x = temp; its better not to use a tautology in your example.
h
heavyd

Prefix notation will increment the variable BEFORE the expression is evaluated. Postfix notation will increment AFTER the expression evaluation.

However "=" has a lower operator precedence than "++".

So x=x++; should evaluate as follows

x prepared for assignment (evaluated) x incremented Previous value of x assigned to x.


This is the best answer. Some markup would have helped it stand out a bit more.
This is wrong. It's not about precedence. ++ has higher precedence than = in C and C++, but the statement is undefined in those languages.
The original question is about Java
C
Community

None of the answers where quite spot on, so here goes:

When you're writing int x = x++, you're not assigning x to be itself at the new value, you're assigning x to be the return value of the x++ expression. Which happens to be the original value of x, as hinted in Colin Cochrane's answer .

For fun, test the following code:

public class Autoincrement {
        public static void main(String[] args) {
                int x = 0;
                System.out.println(x++);
                System.out.println(x);
        }
}

The result will be

0
1

The return value of the expression is the initial value of x, which is zero. But later on, when reading the value of x, we receive the updated value , that is one.


I'll try to understand the bytecode lines, see my update, so it'll be clear.. :)
Using println() was very helpful to me in understanding this.
p
plodoc

It has been already explained well by other. I just include the links to the relevant Java specification sections.

x = x++ is an expression. Java will follow the evaluation order. It will first evaluate the expression x++, which will increment x and set result value to the previous value of x. Then it will assign the expression result to the variable x. At the end, x is back at its previous value.


+1. This is by far the best answer to the actual question, "Why?"
c
cletus

This statement:

x = x++;

evaluates like this:

Push x onto the stack; Increment x; Pop x from the stack.

So the value is unchanged. Compare that to:

x = ++x;

which evaluates as:

Increment x; Push x onto the stack; Pop x from the stack.

What you want is:

while (x < 3) {
  x++;
  System.out.println(x);
}

Definitely the correct implementation, but the question is 'why?'.
The original code was using post-increment on x and then assigning it to x. x will be bound to x before increment, therefore it will never change values.
@cletus I am not the downvoter, but your initial answer didn't contain the explanation. It just said do 'x++`.
@cletus: I didn't downvote, but your answer originally was just the x++ code snippet.
The explanation is incorrect too. If the code first assigned x to x and then incremented x, it would work fine. Just change x++; in your solution to x=x; x++; and you're doing what you claim the original code is doing.
P
Peter Mortensen

The answer is pretty straightforward. It has to do with the order things are evaluated. x++ returns the value x then increments x.

Consequently, the value of the expression x++ is 0. So you are assigning x=0 each time in the loop. Certainly x++ increments this value, but that happens before the assignment.


Wow, there's so much detail on this page when the answer is short and simple i.e. this one.
C
Colin Cochrane

From http://download.oracle.com/javase/tutorial/java/nutsandbolts/op1.html

The increment/decrement operators can be applied before (prefix) or after (postfix) the operand. The code result++; and ++result; will both end in result being incremented by one. The only difference is that the prefix version (++result) evaluates to the incremented value, whereas the postfix version (result++) evaluates to the original value. If you are just performing a simple increment/decrement, it doesn't really matter which version you choose. But if you use this operator in part of a larger expression, the one that you choose may make a significant difference.

To illustrate, try the following:

    int x = 0;
    int y = 0;
    y = x++;
    System.out.println(x);
    System.out.println(y);

Which will print 1 and 0.


It's not the evaluation result that's the issue, though, it's the order of the stores.
I disagree. If x = 0 then x++ will return 0. Therefore x = x++ will result in x = 0.
Rup is right about this. It's the order of the stores which is at issue in this particular case. y=x++ isn't the same as x=x++; On the latter one, x is being assigned 2 values in the same expression. Left hand x is being assigned the result of the evaluation of the expression x++, which is 0. Right hand side x is being incremented to 1. In which order these 2 assignment occur is what the issue is about. From previous posts it is clear that the way this works is: eval = x++ => eval == 0 : increment right x => x == 1 : left x = eval => x == 0
P
Peter Mortensen

You don't really need the machine code to understand what's happending.

According the definitions:

The assignment operator evaluates the right-hand side expression, and stores it in a temporary variable. 1.1. The current value of x is copied into this temporary variable 1.2. x is incremented now. The temporary variable is then copied into the left-hand side of the expression, which is x by chance! So that's why the old value of x is again copied into itself.

It is pretty simple.


P
Peter Mortensen

You're effectively getting the following behavior.

grab the value of x (which is 0) as "the result" of the right side increment the value of x (so x is now 1) assign the result of the right side (which was saved as 0) to x (x is now 0)

The idea being that the post-increment operator (x++) increments that variable in question AFTER returning its value for use in the equation it's used in.

Edit: Adding a slight bit because of the comment. Consider it like the following.

x = 1;        // x == 1
x = x++ * 5;
              // First, the right hand side of the equation is evaluated.
  ==>  x = 1 * 5;    
              // x == 2 at this point, as it "gave" the equation its value of 1
              // and then gets incremented by 1 to 2.
  ==>  x = 5;
              // And then that RightHandSide value is assigned to 
              // the LeftHandSide variable, leaving x with the value of 5.

OK, but what specifies the order of steps 2 and 3?
@Rup - The language defines it. The right side of the equation is evaluated first (in this case, "x++"), and the result is assigned to the variable on the left side. That's how the language works. As far as the "x++" "returning" x for the equation, that's how the postfix increment operator works (return the value of x, then increment it). If it had been "--x", then it would have been (increment x, then return the value). Return isn't the right word there, but you get the idea.
P
Peter Mortensen

This is because it never gets incremented in this case. x++ will use the value of it first before incrementing like on this case it will be like:

x = 0;

But if you do ++x; this will increase.


if you make a test, you can see that it first increments, and later attributes. So it should not attribute zero.
@Tom: see my answer - I show in a test that x++ actually returns the old value of x. That's where it breaks.
"if you make a test" -- some people seem to think that a test written in C tells us what Java will do, when it won't even tell us what C will do.
P
Progman

The value stays at 0 because the value of x++ is 0. In this case it doesn't matter if the value of x is increased or not, the assignment x=0 is executed. This will overwrite the temporary incremented value of x (which was 1 for a "very short time").


But x++ is a post operation. So x would have to be incremented after the assignment is complete.
@Sagar V: only for the expression x++, not for the whole assignment x=x++;
No, I think it only needs to be incremented after the value of x to be used in the assignment was read.
p
patrick

This works how you expect the other one to. It's the difference between prefix and postfix.

int x = 0; 
while (x < 3)    x = (++x);

j
jhabbott

Think of x++ as a function call that "returns" what X was before the increment (that's why it's called a post-increment).

So the operation order is: 1: cache the value of x before incrementing 2: increment x 3: return the cached value (x before it was incremented) 4: return value is assigned to x


OK, but what specifies the order of steps 3 and 4?
"returns what X was before the increment" is wrong, see my update
In reality steps 3 and 4 are not separate operations - it's not really a function call that returns a value, it just helps to think of it that way. Whenever you have an assignment the right hand side is "evaluated" then the result is assigned to the left hand side, the evaluation result can be thought of as a return value as it helps you to understand the order of operations, but it's not really.
Oops, true. I meant steps 2 and 4 - why does the returned value get stored over the top of the incremented value?
This is part of the definition of an assignment operation, first the right hand side is completely evaluated, then the result is assigned to the left hand side.
S
Steven

When the ++ is on the rhs, the result is returned before the number is incremented. Change to ++x and it would have been fine. Java would have optimised this to perform a single operation (the assignment of x to x) rather than the increment.


m
micdah

Well as far as I can see, the error occurs, due to the assignment overriding the incremented value, with the value prior to incrementation, i.e. it undoes the increment.

Specifically, the "x++" expression, has the value of 'x' prior to increment as opposed to "++x" which has the value of 'x' after incrementation.

If you are interested in investigating the bytecode, we will take a look at the three lines in question:

 7:   iload_1
 8:   iinc    1, 1
11:  istore_1

7: iload_1 # Will put the value of the 2nd local variable on the stack
8: iinc 1,1 # will increment the 2nd local variable with 1, note that it leaves the stack untouched!
9: istore_1 # Will pop the top of stack and save the value of this element to the 2nd local variable
(You can read the effects of each JVM instruction here)

This is why the above code will loop indefinitely, whereas the version with ++x will not. The bytecode for ++x should look quite different, as far as I remember from the 1.3 Java compiler I wrote a little over a year ago, the bytecode should go something like this:

iinc 1,1
iload_1
istore_1

So just swapping the two first lines, changes the semantics so that the value left on the top of stack, after the increment (i.e. the 'value' of the expression) is the value after the increment.


P
Paul
    x++
=: (x = x + 1) - 1

So:

   x = x++;
=> x = ((x = x + 1) - 1)
=> x = ((x + 1) - 1)
=> x = x; // Doesn't modify x!

Whereas

   ++x
=: x = x + 1

So:

   x = ++x;
=> x = (x = x + 1)
=> x = x + 1; // Increments x

Of course the end result is the same as just x++; or ++x; on a line by itself.


P
Praveen Prasad
 x = x++; (increment is overriden by = )

because of above statement x never reaches 3;


J
Jay

I wonder if there's anything in the Java spec that precisely defines the behavior of this. (The obviously implication of that statement being that I'm too lazy to check.)

Note from Tom's bytecode, the key lines are 7, 8 and 11. Line 7 loads x into the computation stack. Line 8 increments x. Line 11 stores the value from the stack back to x. In normal cases where you are not assigning values back to themselves, I don't think there would be any reason why you couldn't load, store, then increment. You would get the same result.

Like, suppose you had a more normal case where you wrote something like: z=(x++)+(y++);

Whether it said (pseudocode to skip technicalities)

load x
increment x
add y
increment y
store x+y to z

or

load x
add y
store x+y to z
increment x
increment y

should be irrelevant. Either implementation should be valid, I would think.

I'd be extremely cautious about writing code that depends on this behavior. It looks very implementation-dependent, between-the-cracks-in-the-specs to me. The only time it would make a difference is if you did something crazy, like the example here, or if you had two threads running and were dependent on the order of evaluation within the expression.


c
cubob

I think because in Java ++ has a higher precedence than = (assignment)...Does it? Look at http://www.cs.uwf.edu/~eelsheik/cop2253/resources/op_precedence.html...

The same way if you write x=x+1...+ has a higher precedence than = (assignment)


It's not a question of precedence. ++ has higher precedence than = in C and C++ too, but the statement is undefined.
t
tia

The x++ expression evaluates to x. The ++ part affect the value after the evaluation, not after the statement. so x = x++ is effectively translated into

int y = x; // evaluation
x = x + 1; // increment part
x = y; // assignment

P
Peter Mortensen

Before incrementing the value by one, the value is assigned to the variable.


P
Peter Mortensen

It's happening because it's post incremented. It means that the variable is incremented after the expression is evaluated.

int x = 9;
int y = x++;

x is now 10, but y is 9, the value of x before it was incremented.

See more in Definition of Post Increment.


Your x/y example is different from the real code, and the difference is relevant. Your link doesn't even mention Java. For two of the languages it does mention, the statement in the question is undefined.
p
prime

Check the below code,

    int x=0;
    int temp=x++;
    System.out.println("temp = "+temp);
    x = temp;
    System.out.println("x = "+x);

the output will be,

temp = 0
x = 0

post increment means increment the value and return the value before the increment. That is why the value temp is 0. So what if temp = i and this is in a loop (except for the first line of code). just like in the question !!!!


S
SIVAKUMAR.J

The increment operator is applied to the same variable as you are assigning to. That's asking for trouble. I am sure that you can see the value of your x variable while running this program.... that's should make it clear why the loop never ends.