Tricky stuff

[JosAH]

As everybody knows (or should know) the following is useless:

private static int x= 42;
...
x= x++;
 

The value of x will stay 42. Or does it? Have a look at the following little experiment
and try to figure out what is happening:

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}

kind regards,

Jos

[prometheuzz]

It's because 42 is a special kind of integer, of course!
; )

Here's another one:

public class TrickyStuff {
 
    public static void main(String[] args) {
 
        double a = 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1;
        double b = 0.8;
 
        System.out.println("Are a and b the same? "+(a==b));
    }
}

The output is "Are a and b the same? false". How is this possible?
It's not for you Jos: I know you know the answer.

[JosAH]

It's not for you Jos: I know you know the answer.

I don't want to participate in any contest that wants me to participate :-P

kind regards,

Jos (duh ;-)

[Tripmaker]

It's because 42 is a special kind of integer, of course!
; )

Here's another one:

public class TrickyStuff {
 
    public static void main(String[] args) {
 
        double a = 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1;
        double b = 0.8;
 
        System.out.println("Are a and b the same? "+(a==b));
    }
}

The output is "Are a and b the same? false". How is this possible?
It's not for you Jos: I know you know the answer.

I'm new at programming so I'm going to take a calculated guess. The answer for a = 0.7999999999999999 . The reason being double belongs to Floating-point numbers which represent numbers with fractional precision. Double doesn't round off a number.

Petra

[JosAH]

I'm new at programming so I'm going to take a calculated guess. The answer for a = 0.7999999999999999 . The reason being double belongs to Floating-point numbers which represent numbers with fractional precision. Double doesn't round off a number.

Petra

Very close; bookmark this document and read it when you find some spare time.

kind regards,

Jos

ps. now what about that x == 42 thingie in my original post?

[sumittyagi]

As everybody knows (or should know) the following is useless:

private static int x= 42;
...
x= x++;
 

The value of x will stay 42. Or does it? Have a look at the following little experiment
and try to figure out what is happening:

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}

kind regards,

Jos

Why is it useless?
x is not declared final, and thus is not a constant.

It's because 42 is a special kind of integer, of course!
; )

I didn't get this line. It would be great if you could explain this one.
Thanks!

Regards,
Sumit Tyagi.

[JosAH]

Why is it useless?
x is not declared final, and thus is not a constant.

So you're not surprised by the output it shows?

I didn't get this line. It would be great if you could explain this one.
Thanks!

42 is 54 using base 13.

kind regards,

Jos

[r035198x]

Why is it useless?
x is not declared final, and thus is not a constant.



I didn't get this line. It would be great if you could explain this one.
Thanks!

Regards,
Sumit Tyagi.

It's not a constant alright but when you do

while(true) {
      x = x++;
}

the value of x does not change since x will be assigned to the previous value always which will be the initial value. The post increment won't take any effect there. But when a thread is reading off the values of x ...

[Tripmaker]

As everybody knows (or should know) the following is useless:

private static int x= 42;
...
x= x++;
 

The value of x will stay 42. Or does it? Have a look at the following little experiment
and try to figure out what is happening:

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}

kind regards,

Jos

Let see if I can get this right. The value stays at 42 because of the while statement. In laymans terms while x == 42 the system prints out x != 42 and the program is ended. Everytime you re-run the program you keep getting the same answer. You will never get to the second while statement.

[r035198x]

Let see if I can get this right. The value stays at 42 because of the while statement. In laymans terms while x == 42 the system prints out x != 42 and the program is ended. Everytime you re-run the program you keep getting the same answer. You will never get to the second while statement.

Put the program on your compiler and test it. Realize also that there are two threads running here.

[sumittyagi]

So you're not surprised by the output it shows?

Yea! I am not surprized by the output, because output is as I expected it to be.
---------- java ----------
x= 42
x != 42
Normal Termination
Output completed (1 sec consumed).

42 is 54 using base 13.

But how does that make a difference. I mean does this fact have any influence on the output.

[r035198x]

Yea! I am not surprized by the output, because output is as I expected it to be.
...

Would you then care to explain why the value of x changes (or seems to change)?

[sumittyagi]

It's not a constant alright but when you do

while(true) {
      x = x++;
}

the value of x does not change since x will be assigned to the previous value always which will be the initial value. The post increment won't take any effect there. But when a thread is reading off the values of x ...

x++ is having a side effect of incrementing the value of x.
so it will be solved as follows:
x = x++;
x = 42 (value replaced, so x becomes 42);
x++ (x++ have side effect of incrementing x, so now x becomes 43);

[prometheuzz]

Yea! I am not surprized by the output, because output is as I expected it to be.
---------- java ----------
x= 42
x != 42
Normal Termination
Output completed (1 sec consumed).

Huh? Have we compiled and run the same code (from the original post)? I don't get that output!

But how does that make a difference. I mean does this fact have any influence on the output.

That was a joke of me (42 being a special kind of integer).

[sumittyagi]

Huh? Have we compiled and run the same code (from the original post)? I don't get that output!

Then what output are u getting?

That was a joke of me (42 being a special kind of integer).

There are a lots of things in java that make you bite your finger, so I never take any statement as joke.

But that was nice one!!! LOL.

[r035198x]

Then what output are u getting?



There are a lots of things in java that make you bite your finger, so I never take any statement as joke.

But that was nice one!!! LOL.

I'm getting Sumit's output and I don't think the output is platform dependant (or is it?).

[sumittyagi]

42 is 54 using base 13.

JosAH! you are having very good sense of humor.
LOL.

[prometheuzz]

@sumittyagi & r035198x:

Now I'm confused...
The output I get is this: x= 42, and the application keeps running, which was what I had expected. I though that was what Jos had meant by this: the fact that the line of code "System.out.println("x= "+x);" would or wouldn't be executed. Because the lines of code before that println() gets executed in it's own thread, "System.out.println("x= "+x);" gets executed, of course. I didnt think much of it, and I didn't find it tricky at all!

But I would have been surprised if x != 42 would be printed as well. I've compiled and run the code with Eclipse, which does not use Sun's javac, and with Sun's javac. Both of them produce the same output: x= 42.

[prometheuzz]

... and I don't think the output is platform dependant (or is it?).

No, I can't imagine that.

[r035198x]

@sumittyagi & r035198x:

Now I'm confused...
The output I get is this: x= 42, and the application keeps running, which was what I had expected. I though that was what Jos had meant by this: the fact that the line of code "System.out.println("x= "+x);" would or wouldn't be executed. Because the lines of code before that println() gets executed in it's own thread, "System.out.println("x= "+x);" gets executed, of course. I didnt think much of it, and I didn't find it tricky at all!

But I would have been surprised if x != 42 would be printed as well. I've compiled and run the code with Eclipse, which does not use Sun's javac, and with Sun's javac. Both of them produce the same output: x= 42.

I get x != 42 printed alright (vista and jdk 1.5)Which OS are you using?
I thinks that's why Jos is actually questioning whether x stays as 42 or not. Because x actually appears to change in this one and the program terminates.

[sumittyagi]

@sumittyagi & r035198x:

Now I'm confused...
The output I get is this: x= 42, and the application keeps running, which was what I had expected. I though that was what Jos had meant by this: the fact that the line of code "System.out.println("x= "+x);" would or wouldn't be executed. Because the lines of code before that println() gets executed in it's own thread, "System.out.println("x= "+x);" gets executed, of course. I didnt think much of it, and I didn't find it tricky at all!

But I would have been surprised if x != 42 would be printed as well. I've compiled and run the code with Eclipse, which does not use Sun's javac, and with Sun's javac. Both of them produce the same output: x= 42.

I really don't know how u r getting that output. that is really strange.
Now its out of my knowledgebase.
so one thing we can do is....

SOS... signals to josAH!!!!

[sumittyagi]

I really don't know how u r getting that output. that is really strange.
Now its out of my knowledgebase.
so one thing we can do is....

SOS... signals to josAH!!!!

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}
 

the kind of output you are saying is only possible in case when control doesn't leave main thread till main thread vanishes (i.e non-preemptive scheduling at OS level).

try executing this code, and tell us what the output u r getting.

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        while(true) {
           x=x++;
           System.out.println("x= "+x);
        }
    }
}
 

[sumittyagi]

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}
 

the kind of output you are saying is only possible in case when control doesn't leave main thread till main thread vanishes (i.e non-preemptive scheduling at OS level).

try executing this code, and tell us what the output u r getting.

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        while(true) {
           x=x++;
           System.out.println("x= "+x);
        }
    }
}
 

Sorry! I think I misunderstood the concept, and went on saying according to that. I think the thing I said about non-preemptive scheduling is not the cause.

And now I got it, why josAH said it is tricky stuff.
try this code.

public class TrickyStuff  {
 
    private static volatile int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}
 

I think now you should get the same output as we are getting.

[blazedaces]

This is my output:

--------------------Configuration: <Default>--------------------
x= 42
x != 42
 
Process completed.

I'm still confused though, why is this tricky? Shouldn't ++ following a variable tell the compiler do this variable = variable + 1 only after you complete the statement on this line? So wouldn't x = x++; tell the compiler ok, x = x, now that you're done, do x = x + 1?

Is this not exactly what happens? I would be surprised if it didn't happen and I'm getting the impression that there are situations where that is the case from you guys?

Interesting topic anyway...

-blazed

[prometheuzz]

Sorry! I think I misunderstood the concept, and went on saying according to that. I think the thing I said about non-preemptive scheduling is not the cause.

And now I got it, why josAH said it is tricky stuff.
try this code.

public class TrickyStuff  {
 
    private static volatile int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}
 

I think now you should get the same output as we are getting.

Nope, still the same: only x= 42 is printed to the console. Both Eclipse and Sun's compiler do this. Also, before recompiling with javac, I removed all .class files, just to be sure.

[prometheuzz]

I get x != 42 printed alright (vista and jdk 1.5)Which OS are you using? ...

I can't imagine it's the OS, but here's what I use:
Windows XP
Eclipse SDK 3.2.0
Sun's JDK 1.6

I'll also try my Linux box when I get home.

[r035198x]

I can't imagine it's the OS, but here's what I use:
Windows XP
Eclipse SDK 3.2.0
Sun's JDK 1.6

I'll also try my Linux box when I get home.

Bart, look at Jos' thread again (I mean Java thread), when started it simply does

System.out.println( "x != 42" );

His while is of no importance because he has a semi colon after it. So as long as that thread starts, the program exits printing x != 42 when x actually is 42.

[prometheuzz]

...
His while is of no importance because he has a semi colon after it.
...

I don't see why it is of no importance. It keeps looping on that while in the java.lang.Thread.
When I execute this code:

public class TrickyStuff {
 
    private static int x = 42;
 
    public static void main(String[] args) {
        (new Thread() {
            public void run() {
                while(x == 42){ System.out.println("x is the same"); }
                System.out.println("x != 42");
                System.exit(0);
            }
        }).start();
 
        System.out.println("x = "+x);
 
        while(true) { x = x++; }
    }
}

First x = 42 is printed and then x is the same keeps being printed to the console. It never reaches x != 42. At least that's what's happening here. Really.

[r035198x]

I don't see why it is of no importance. It keeps looping on that while in the java.lang.Thread.
When I execute this code:

public class TrickyStuff {
 
    private static int x = 42;
 
    public static void main(String[] args) {
        (new Thread() {
            public void run() {
                while(x == 42){ System.out.println("x is the same"); }
                System.out.println("x != 42");
                System.exit(0);
            }
        }).start();
 
        System.out.println("x = "+x);
 
        while(true) { x = x++; }
    }
}

First x = 42 is printed and then x is the same keeps being printed to the console. It never reaches x != 42. At least that's what's happening here. Really.

It's definitely platform dependant then. When I run that version I get different results each time.
The first statements are either "x = 42"
or "x is the same" with "x is the same" repeated sometimes then it prints "x != 42" and exits.


I've got "x = 42" printed after "x is the same" sometimes (but I have it printed only once both times). Your output assumes that Jos' thread is the only thread that getting executed once it starts.

[nomad]

As everybody knows (or should know) the following is useless:

private static int x= 42;
...
x= x++;
 

The value of x will stay 42. Or does it? Have a look at the following little experiment
and try to figure out what is happening:

public class TrickyStuff {
 
 
public static void main(String[] args) {
 
(new Thread() {
public void run() {
while(x = 42);
System.out.println( "x != 42" );
System.exit(0);
}
} ).start();
 
System.out.println("x= "+x);
while(true) { x=x++;}
}
}

kind regards,

Jos

System.out.println( "x != 42" ); just prints x!=42 it has no value other than printing out text.

while(true) { x=x++;} is skiped durning the run.

nomad

[JosAH]

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}

When I run this code (Eclipse 3.2 Java 1.6.0_01) I get as output:

x= 42
x != 42
 

and the process never stops; the other thread has terminated but the main
thread keeps on running. When I change the 'while (true)' thing in the main
thread to, say:

boolean flag= true;
while (flag) ...
 

I only get the output 'x= 42'. It is tricky stuff; when the compiler compiles whithout
optimizing anything x does become 43 for a very short moment and after the
old value (42) is stored in x again. If in that very short moment the other thread
tests for the value of 'x' it can terminate its while loop. The main thread never
terminates though.

It all depends on thread scheduling and when the values of 'x' are written back
to main memory where all (both) threads can access it. It is tricky stuff indeed ;-)

kind regards,

Jos

[blazedaces]

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}

When I run this code (Eclipse 3.2 Java 1.6.0_01) I get as output:

x= 42
x != 42
 

and the process never stops; the other thread has terminated but the main
thread keeps on running. When I change the 'while (true)' thing in the main
thread to, say:

boolean flag= true;
while (flag) ...
 

I only get the output 'x= 42'. It is tricky stuff; when the compiler compiles whithout
optimizing anything x does become 43 for a very short moment and after the
old value (42) is stored in x again. If in that very short moment the other thread
tests for the value of 'x' it can terminate its while loop. The main thread never
terminates though.

It all depends on thread scheduling and when the values of 'x' are written back
to main memory where all (both) threads can access it. It is tricky stuff indeed ;-)

kind regards,

Jos

Wow, really cool to know. Good stuff man...

-blazed

[r035198x]

public class TrickyStuff  {
 
    private static int x= 42;
 
    public static void main(String[] args) {
 
        (new Thread() {
            public void run() {
                while(x == 42);
                System.out.println( "x != 42" );
                System.exit(0);
            }
        } ).start();
 
        System.out.println("x= "+x);
        while(true) { x=x++;}
    }
}

When I run this code (Eclipse 3.2 Java 1.6.0_01) I get as output:

x= 42
x != 42
 

and the process never stops; the other thread has terminated but the main
thread keeps on running. When I change the 'while (true)' thing in the main
thread to, say:

boolean flag= true;
while (flag) ...
 

I only get the output 'x= 42'. It is tricky stuff; when the compiler compiles whithout
optimizing anything x does become 43 for a very short moment and after the
old value (42) is stored in x again. If in that very short moment the other thread
tests for the value of 'x' it can terminate its while loop. The main thread never
terminates though.

It all depends on thread scheduling and when the values of 'x' are written back
to main memory where all (both) threads can access it. It is tricky stuff indeed ;-)

kind regards,

Jos

It has to be a very short period of time indeed because when I put

while(x == 42);
                System.out.println(x+ " x != 42" );

it always printed "x = 42 x != 42".
Your sleeves are tricky place to be Jos.

[JosAH]

It has to be a very short period of time indeed because when I put

while(x == 42);
                System.out.println(x+ " x != 42" );

it always printed "x = 42 x != 42".
Your sleeves are tricky place to be Jos.

Could very well be: the System.out Stream is shared too between the two threads.
I have never seen that output but it isn't impossble. You stated that your process
terminates in a previous reply; that's the part I don't understand: the main thread
should loop until hell freezes over ;-)

kind regards,

Jos

[JosAH]

Your sleeves are tricky place to be Jos.

ps.

One should always explore the darkest corners of a (programming) language to
really get to know the language ;-)

kind regards,

Jos

[r035198x]

...You stated that your process
terminates in a previous reply; that's the part I don't understand: the main thread
should loop until hell freezes over ;-)

kind regards,

Jos

Isn't it that after "x != 42" is printed then System.exit(0); terminates the program?

[JosAH]

Isn't it that after "x != 42" is printed then System.exit(0); terminates the program?

Ah, yes, of course (I'm a bit tired), I was wrong of course; forget that remark.
What also surprises me is that Prometheuzz doesn't get that output; tricky stuff ...

kind regards,

Jos

[r035198x]

Ah, yes, of course (I'm a bit tired), I was wrong of course; forget that remark.
What also surprises me is that Prometheuzz doesn't get that output; tricky stuff ...

kind regards,

Jos

There's quite a few tricky areas in Java too. Remember that JFrame named Ghost?

[JosAH]

There's quite a few tricky areas in Java too. Remember that JFrame named Ghost?

Erm, nope; enlighten me please; I like tricky things ;-)

kind regards,

Jos

[r035198x]

Erm, nope; enlighten me please; I like tricky things ;-)

kind regards,

Jos

If you call a JFrame or awt Frame "Ghost" on some windows platforms (windows XP is one of them) it wont be displayed even if you do setVisible(true);

[r035198x]

If you call a JFrame or awt Frame "Ghost" on some windows platforms (windows XP is one of them) it wont be displayed even if you do setVisible(true);

Something like

import javax.swing.*;
public class Test {
  private static JFrame ghost = new Ghost(); 
 
  public static void main(String[] args) {
      ghost.setVisible(true);
      System.out.print("You can't see me");
  }
 
}
 
class Ghost extends JFrame { }
 

Won't show on both 1.4 and 1.5 on XP

[JosAH]

Something like

import javax.swing.*;
public class Test {
  private static JFrame ghost = new Ghost(); 
 
  public static void main(String[] args) {
      ghost.setVisible(true);
      System.out.print("You can't see me");
  }
 
}
 
class Ghost extends JFrame { }
 

Won't show on both 1.4 and 1.5 on XP

Cute! I have Java 1.6 and the window does show; tomorrow at the office I'll try it
on a previous Java version. I like those Easter Eggs ;-)

kind regards,

Jos

[r035198x]

Cute! I have Java 1.6 and the window does show; tomorrow at the office I'll try it
on a previous Java version. I like those Easter Eggs ;-)

kind regards,

Jos

It's always good to see the program disobey you for once. I'm tempted to stick this so people can people can keep the treats coming.

[sumittyagi]

It's always good to see the program disobey you for once. I'm tempted to stick this so people can people can keep the treats coming.

Great!!!!!
It was a yummy thread!!!! So many surprizing things.
I enjoyed it a lot.