Ticklish Techs

In fact one OpenFileDialog is enough but the .net framework provides two of them:

1. System.Windows.Forms.OpenFileDialog in System.Windows.Forms.dll
2. Microsoft.Win32.OpenFileDialog in PresentationFramework.dll (that is WPF)

With such a setup one would use the first for WinForms applications and the second for WPF apps. In fact if you create a WPF application and just type in OpenFileDialog you will get number 2.

But the 2nd one (even if it is part of the newer part of the framework) is an old dialog where the 1st will use the new and shiny Windows 7 dialog on Win7. Just look at the following screenshots.

Only the first class uses the Windows 7 dialog and comes with proper support for Libraries etc. Even if the second shows the Libraries in its left part it does not fully support this new Windows 7 feature.

I have no idea why the WPF class uses the old windows API but it does. For best compatibility with all Windows versions (including 7) you should always use the Winforms Dialog even when building WPF apps! It does not hurt to reference the System.Windows.Forms.dll (it is installed an every machine running the .net framework) and you can use that dialog from WPF without any trouble.

You should only know about one little disadvantage when using the WinForms dialog. It only accepts WinForms as a parent window no WPF windows. To overcome this limitation please read my other article about the IWin32Window interface and how to implement / provide it in WPF at http://www.ticklishtechs.net/2009/12/22/how-to-get-a-iwin32window-from-a-wpf-window/

Disclaimer: The following is no official advice or anything. I just did it and it worked ON MY MACHINE. Fell free to try but have a backup of your data.

The Windows 7 RTM bits will not allow an upgrade from an earlier version of Windows 7 (e.g. RC or beta). I belief that Microsoft did this to ensure that your new operating system works just fine and no beta and release bits get mixed up.

For the upgrade from the beta version to the release candidate quickly some patch occurred which allowed an upgrade. I used the same one and it works fine on my machine to upgrade Windows 7 RC to RTM.

What’s to do?

You need to change one file on the installation DVD. On way to do this is to copy all files to your computer, change the file and burn a new DVD. Since the upgrade will be run from within the already installed Windows 7 you do not need to boot from this DVD. I just copied all files to a USB drive and run the setup.exe from that drive. I belief you will be able to install Windows also from a directory on you local hard disk containing all the files from the DVD.

In the source folder you will find a text file named cversion.ini. The original file looks like:

[HostBuild]
MinClient=7233.0
MinServer=7100.0

Just change the MinClient number to the build version of you current Windows version. For the Windows 7 RC it will be 7100.

The upgrade should work fine now. But if your new Windows 7 starts acting strange don’t blame Microsoft or me. Always remember you installed it with an unsupported patch.

Here is a nice little challenge, Benjamin came up with some months ago: “From a list of numbers find the smallest and the largest one – but do not use more than 1,5 comparisons per number.”

Dear reader: Try it!

The solution is simple, straight forward and very nice. And I didn’t find it.

I found something else. My approach was to look at the largest-number-so-far (max) and the smallest-number-so-far (min) as the boundaries of a region. A number outside this region must be larger than max or smaller than min and causes that boundary to change.

To check if a number (a) was outside that region a little computation came to my mind:

check = (max - a)*(min - a);

You would expect n to be smaller than max, so (max-n) should be positive.

You would expect n to be larger than min, so (n-min) should be positive, too.

Two positive numbers multiplied result in a positive number again.

So, if n was outside the boundaries, check would become negative. I only had to check which boundary was exceeded and that’s it:

if (check < 0)
    if (a > max)
        max = a;
    else
        min = a;

Using this approach the number of comparisons  converges to 1 when the length of the list grows. So I found a solution that is way better than 1,5 comparisons per number, didn’t I?

Boooooh

No, I didn’t. I cheated. In fact  (max-a) and (min-a) are comparisons, they just don’t use > or <.

(Actually it’s the other way around: To compute < or > most processors do a subtraction and compare the result to 0.)

So – if you count the subtractions as well you get 3+ comparisons per number…

The intended solution to the challenge (and the code you should provide in your exam) is:

a = list[count++];
b = list[count++];

if (a<b)
{
    if (a<min) min = a;
    if (b>max) max = b;
}
else
{
    if (b < min) min = b;
    if (a > max) max = a;
}

So – what’s the point of this post?

The point is: My silly approach can be faster than the standard solution. Depending on the type and range of the numbers in the list calculating and probing check needs less time than the comparisons in the standard solution.

It may not be much, but sometimes small advantages matter.  For example: For an integer-list of length 10^8 with values from -10.000 to 10.000 my approach is 0.02 seconds faster (on my current laptop). And has less code.

So if you feel like using it – I won’t charge you.

Disclaimer: I’m absolutely not the person who can tell a lot about Java. I learned it a few years ago for university, but used it most of the time only for small example stuff. Never as deep as I’m working with .net. But maybe my newest discoveries are useful for more programmers with .net background.

I started using Java 1.4 and until last week I never took a look at Java 5 or 6. The only thing I was aware of were generics in Java.

iterators in Java

I was always annoyed in the missing of a foreach loop in Java. There is a concept of iterators in Java comparable to the .net Enumerators. But to loop over a list you will find in millions of Java programs the following code:

Iterator<Integer> i = list.iterator();
while(i.hasNext()) {
    Integer current = i.next();

    // do something with the current element
}

Sometimes build into a regular for loop, but I don’t think that is more readable or convenient:

for(Iterator<Integer> i = list.iterator(); i.hasNext(); ) {
    Integer current = i.next();

    // just do something
}

By the way: Can anyone tell me why the method is called iterator() and not getIterator() as used everywhere else in Java?

foreach in Java

With Java 5 (or 1.5 – call it as you like) a foreach loop was introduced. This is about five years ago, so I’m very fast in getting news from the Java world.

for(Integer i : list) {
    // just use i as the current element
}

I like it. It’s working for everything that implements java.lang.Iterable and for regular arrays.

Imaging a small C# winform app with just one ListBox and the following code:

private void Form1_Load(object sender, EventArgs e)
{
    for (int i = 0; i < 5; i++)
    {
        Thread t = new Thread(() => AddToListbox(i));
        t.Start();
    }
}

private void AddToListbox(int i)
{
    if (this.InvokeRequired)
        this.Invoke(new Action<int>(AddToListbox), i);
    else
        this.listBox1.Items.Add(i);
}

A simple loop iterating over the numbers from 0 to 4 and adding these values asynchronous to a ListBox. What do you expect? I expected the numbers from 0 to 4 shown in the ListBox but in any random order since I do not control the threads in any way.

I didn’t expect any number to appear multiple times and I’m totally surprised to  see the number 5!

But ReSharper gave me a hint that I often saw but never understood:

So what’s going on?

I use the syntax of an Lambda expression instead of a regular function call (e.g. using the ThreadStart class and a delegate). This Lambda expression is not evaluated until the thread uses it. And by this time, the loop can be in its next iteration. If the loop is already finished i will be 5.

That is exactly what R# tries to tell me: “Hey, you are accessing here a variable but change it later on. Maybe that is not a good idea.”. – It isn’t.

The solution

Just make a copy of i before passing it into the expression. This copy must be a private copy that will not be changed later. The easiest way to do so is declaring a variable inside the body of the loop. In every iteration of the loop a new integer will be created on the stack and the Lambda expression will access this one.

for (int i = 0; i < 5; i++)
{
    int copy = i;
    Thread t = new Thread(() => AddToListbox(copy));
    t.Start();
}

Ten days ago I promised a proof why the differences of any permutation of the same digits is always a multiple of 9. Here it is

Let’s say x and y are digits and ‘xy’ is not ‘x*y’, but the number that consist of the digits x and y.

xy – yx = multiple of 9.

Is it always true? Yes, it is:

(10x + 1y) – (10y + 1x) = multiple of 9?
9x – 9y = multiple of 9?
9(x-y) = multiple of 9? Yes, for sure.

I calculated the numeric values of xy and yx by "10 times the higher position + 1 time the lower position", just like we all do every day in our beloved decimal-system. More formally we describe the factor for the position (1, 10, 100…) by

10^position, (position is zero-based, counting from right to left, of course)

Now, when we change the position of a single digit within a number we change its numeric value from

x*10^(old position) to x*10^(new position).

We can neglect x here, because it’s a factor that occurs in both values, so we can get rid of it by division.
The remaining part for building the difference is

10^p1 – 10^p2, a formula that always produces multiples of 9: 10-1, 1000-100, 10-1000000.

Conclusion

Since the difference of a number and one of it’s permutations is a sum of (10^p1-10^p2)-parts, which are all multiples of 9, the total also can be divided by 9.

Lately I stumbled over permutations and noticed a funny fact: When you take any integer, produce a permutation and subtract one form the other, you always get a multiple of 9.

It works with a any length. Some examples:

Length 2
23 – 32 = -9 
84 – 48 = 36
60 – 06 = 54

Length 3
123 – 132 = -9
and so on…

It still works for changing the front position:
123 – 213 = -90

In fact it works for changing any position:
123 – 321 = -198 (-22*9)

This leads to the conclusion I stated in the beginning. The difference of any permutation of the same number is a multiple of 9. Feel free to check it with you favorite numbers.

It’s not a great mathematical invention (if it is, please let me know), but I didn’t know it, didn’t expect it und find it kind of fascinating. I asked a couple of friends and no one was aware of this.

I still try to find a way to use these fact, maybe for spell-check-like for numbers or a fast calculation of permutations, but so far it seems just useless

Instead I can provide a proof for numbers of any length and I’ll post it here in ten days from today. Meanwhile I want you to try it yourself. The proof is really, really simple, almost trivial and it won’t cost you more than five minutes to understand it all.

Please email or comment your solution!

For those who look for a quick solution:

Try switching off the win32manifest-switch in Visual Studio.

Here is the long story.

Hi! As you may already know we are working on the best .NET – ARIS interface there is so far. It’s an interface that makes heavy use of an old C-Api, using interop and p/invoke.

In Mid-December Benjamin found a very strange behavior that almost spoiled my Christmas. We had just release Version 1.1 when he called me: “Arisan crashes with Vista64.” I was shocked. “And with Vista32 as well”. I felt annihilated. We had done so much testing with various versions of XP, Vista and it worked fine with any of those OSes.

After some testing I saw that ARIS didn’t deliver any pointers at all. That’s rather bad for an interface to C that uses pointer in every call.  After one more day we had more evidence: We had switched to .NET3.5 and VS2008 some month ago and our old versions, that were compiled using the .NET2.0-compiler, still worked.

I was so terrified. The 3.5-framework should be similar to the 2.0-framework, just some (not so) little enhancements. A bug in the 3.5-compiler?

To provide a quick solution we re-wrote parts of the fancy 3.5-syntax, to make our newest enhancement compile with the 2.0-compiler again. That worked (at least we thought so) and we released version 1.1.1.

Then a very frustrating time began. We read articles over and over and quickly found this very interesting post. The problems described there where exactly what we saw.  So we followed the track of the nxcompat-flag and DEP, but it didn’t lead us anywhere.

The C-Api we interface is well documented, but it calls hundreds of other components and one of them crashed deep inside an undiscovered country.

We read pages of differences between 2.0- and 3.5-compiler, new compiler-flags (should have paid more attention here), any documentation we could find, but nothing seemed to help.

We had a feeling it could have something to do with Vista UAC, but we didn’t find any matching explanations to our problem. Being really desperate we started this thread at the MS forums. As you can read there, no one could provide a solution that worked.

I was really sad, because the last thing I wanted to do was to switch the entire development of arisan back to VS2005. I got prepared for a very sad life from now on.

But five days ago my life turned to happiness again. One more time I started of by googling words like “.NET compiler 2.0 3.5 differences” and one more time I read the new features of the 3.5-compiler. This time I obviously paid more attention to the compiler-flags and read about the /win32manifest and /nowin32manifest – switches. I didn’t have much hope left, but tried to set the /nowin32manifest switch.

It worked! Everything worked! The C-Api delivered pointers again. I tried switching the win32manifest on and off and a day later Benjamin confirmed my results. We had found it. But after a long time of suffering (for me it felt like half a year) we still were skeptical and thought we needed a deeper understanding of the problem.

The win32manifest and UAC

Finally we had something to look for and quickly found valuable information on the great I’m just sayin’ blog. The documentation of the switches helped as well, so here is a summary:

The win32manifest is not the .NET-manifest.

The win32manifest is part of the UAC (user access control). UAC was introduced with Vista as an answer to more and more security threads. One of its ideas is to keep malicious software from secretly starting to work. If you work with Vista you may have noticed a lot of dialogs that ask you for permission to run certain programs. THAT is UAC.

Now the win32manifest inside an application tells the OS what level of permission it needs to run. There are levels like “asInvoker” (default), “highestAvailable” or “requireAdministrator”. In case the OS (and the current user) can provide the requested permission, the program runs as a trusted process (according to the granted access level). Of course, 90+% (just my estimation) of the programs on you Vista-machine don’t have a win32mianifest,  because it’s a new feature and older applications just don’t have it. (And I think a majority of newer applications won’t have it either). In order to be compatible with no-win32manifest applications on one hand and providing security on the other hand, Vista can do a miracle called “virtualization”. It’s a bit like running a process in a sandbox: Vista provides anything the process may need to run: disk space, memory, a registry etc. So the process feels cosy.

But anything provided this way is separated from the main system. The process uses a copy of the registry, writes to a special parts of the HDD drives and accesses the memory in a controlled way (not sure about the memory thing…). For me this is a great achievement of Vista, almost magic. I can’t tell in detail how it works (mainly because I don’t have a clue), but it works good and doesn’t cost notable performance.

So on your Vista machine you have processes that run virtualized and others that run non-virtualized. “Non-virtualized” is also called “UCA-compatible” or just “compatible”, so I’ll adopt this wording here.

You could say that Vista trusts compatible processes a little bit more than virtualized processes. That doesn’t mean virtualized processes are evil, but.. well… maybe… under certain circumstances… not 100% trustworthy. (Here is another nice article)

Note three facts here:

1. The state (virtualized/compatible) always counts for a process and is set when the process starts. When you start an .exe-file, Vista decides how to run it.
2. Because of 1) there is no sense in attaching a win32manifest to components that do not start a process, like .dlls. Thus .dlls don’t have a win32manifest.
3. The state (virtualized/compatible) of a process doesn’t change during its lifetime.

So if a .dll is used by a virtualized process, its code also runs virtualized. And if the same .dll is used by a compatible process, its code runs compatible.

Interaction

With some of that in my mind, I could explain the problems we had with our interface software arisan: arisan is a .dll you reference in your .NET-application and use it. As mentioned above arisan calls a native (so non-.NET) .dll using p/invoke-interop-techniques. Now this native .dll starts a bunch of new processes: a database-server, a database-driver, some middle-layers, etc. I am pretty sure non of this newly started processes is based on applications with a win32manifest – so they all run virtualized.

But the .NET-applications we use to test the arisan.dll started running compatible as we switched to VS2008 and the 3.5-compiler. So our test application (compatible) called arisan.dll (compatible), arisan.dll (compatible) called C-Api.dll (compatible). C-Api.dll started new processes (virtualized !!) and called functions there. And in the end a compatible (fully trusted) process asked a virtualized (less trusted) process to fill some pointers… and that’s the point where Vista interferes and says: “Dudes, are you drunk? No way!”

It was a bit of bad luck that the clash of a virtualized and a compatible process happened so deep inside a software we couldn’t debug and only threw some strange, meaningless error-messages, but on the other hand Vista could have told us more. Maybe there is some place in Vista where the clash is logged. If someone knows – please let me know, too. Let me get this straight: Vista is absolutely right here, but I’d wish more information when it happens. I don’t exactly know how, but you could do harmful things if you could easily exchange all sorts of pointers between virtualized and compatible processes…

Solution

So finally the solution was easy. Both processes have either to run virtualized or non-virtualized. Since we can’t make a third-party software run compatible, we have to make our .NET-application run virtualized: We just need to prevent .NET from inserting a win32manifest to executables.

Using the compiler from the command-line the compiler-flag /nowin32manifest does it.

In VS2008 same switch can be found on the “Application”-tab of the project properties.

The manifest-area is enabled for ‘console applications’ and ‘Windows applications’. It is disabled for class libraries.

Another way to handle this problem is to switch off UAC completely. But we strictly advice you against this solution. UAC is one of Vista’s main features and it is designed to provide shelter for your precious data in times of evil viruses and brutal intrusion attempts

Epilog- bad fix

After we understood the problem, we saw that the ‘fix’ we provided as V1.1.1 couldn’t do much. Ok – it made the demos run, because the executable compiled with the 2.0- come without a win32manifest and thus run virtualized. But users trying it with VS2008 still had the same issues.

Actually we cannot provide a technical solution with the arisan.dll, because if the running mode depends on the executables that use arisan. So we’ll write detailed information and everyone lived happily ever after.

Ok, what does this expression do or mean? First of all: never write something like this in your code. Nobody understands the meaning with a single look and that will lead to confusion and maybe to a bug in your software someday.

Now let’s take a deeper look. First you may think about the difference between x++ and ++x. If you write it as a single statement both expressions are equal, but if you use it as a part of a complex expression the first will evaluated to the value of x before increasing it; the second one will evaluate to the new value of x. So y = x++; leads to a different result for y as y = ++x;.

x+=a simply is a ‘shortcut’ for x=x+a.

Now let’s do it step by step. For example x is 5. Then first x will be increased by one to 6 but the old value will go into the formula that remains as x=x+5. Since x was increased before the result will be 11.

If you think that is all right, than please take a break and test it with your favorite compiler. If you are a C or C++ guy you will in fact receive 11 as an answer and everything is fine. But if you are a C# or java guy x will be 10. Why?

.Net as well as the Java Runtime are stack machines. The expression will be put on the stack step by step before evaluating the whole thing. At that time x is 5. x will be changed to 6 by the x++ part, but that only happens in the main memory. The old value (5) is still on the stack. After executing the whole expression the changed x will be overwritten by 10 (5+5).

And once again: NEVER write code like this!

I found this somewhere and I think it’s worth to write a short (only a very short) entry.

throw null;

If you use this expression in your C# code it will throw a NullReferenceException. That is because the throw-statement needs an object of type Exception as its single parameter. But this very object is null in my example.

I like this way to produce some exception while testing code.