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  • C# Convert negative int to 11 bits

    - by Klemenko
    I need to convert numbers in interval [–1024, 1016]. I'm converting to 11 bits like that: string s = Convert.ToString(value, 2); //Convert to binary in a string int[] bits = s.PadLeft(11, '0') // Add 0's from left .Select(c => int.Parse(c.ToString())) // convert each char to int .ToArray(); // Convert IEnumerable from select to Array This works perfectly for signed integers [0, 1016]. But for negative integers I get 32 bits result. Do you have any idea how to convert negative integers to 11 bits array?

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  • Why is TreeSet<T> an internal type in .NET?

    - by Justin Niessner
    So, I was just digging around Reflector trying to find the implementation details of HashSet (out of sheer curiosity based on the answer to another question here) and noticed the following: internal class TreeSet<T> : ICollection<T>, IEnumerable<T>, ICollection, IEnumerable, ISerializable, IDeserializationCallback Without looking too deep into the details, it looks like a Self-Balancing Binary Search Tree. My question is, is there anybody out there with the insight as to why this class is internal? Is it simply because the other collection types use it internally and hide the complexities of a BST from the general masses...or am I way off base?

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  • What's a good algorithm for searching arrays N and M, in order to find elements in N that also exist

    - by GenTiradentes
    I have two arrays, N and M. they are both arbitrarily sized, though N is usually smaller than M. I want to find out what elements in N also exist in M, in the fastest way possible. To give you an example of one possible instance of the program, N is an array 12 units in size, and M is an array 1,000 units in size. I want to find which elements in N also exist in M. (There may not be any matches.) The more parallel the solution, the better. I used to use a hash map for this, but it's not quite as efficient as I'd like it to be. Typing this out, I just thought of running a binary search of M on sizeof(N) independent threads. (Using CUDA) I'll see how this works, though other suggestions are welcome.

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  • Tree + Recursion..

    - by RBA
    Hi.. I came across an article on Binary Trees Search . It uses intensive Recursive Algorithms.. I am just so confused with these stuff.. Please guide my path so as I understand these problems at ease, or any good website to read about recursion first and then solving these problems.. Please share your experience on it.. Its very urgent, and I want to learn these concepts as soon as possible.. Thankss... Regards.

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  • VB.NET encoding one character wrong

    - by Nick Spiers
    I have a byte array that I'm encoding to a string: Private Function GetKey() As String Dim ba() As Byte = {&H47, &H43, &H44, &H53, &H79, &H73, &H74, &H65, &H6D, &H73, &H89, &HA, &H1, &H32, &H31, &H36} Dim strReturn As String = Encoding.ASCII.GetString(ba) Return strReturn End Function Then I write that to a file via IO.File.AppendAllText. If I open that file in 010 Editor (to view the binary data) it displays as this: 47 43 44 53 79 73 74 65 6D 73 3F 0A 01 32 31 36 The original byte array contained 89 at position 11, and the encoded string contains 3F. If I change my encoding to Encoding.Default.GetString, it gives me: 47 43 44 53 79 73 74 65 6D 73 E2 80 B0 0A 01 32 31 36 Any help would be much appreciated!

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  • Dictionary not deserializing

    - by Shadow
    I'm having a problem where one Dictionary in my project is either not serializing or not deserializing. After deserializing, the data I serialized is simply not in the object. Here's the relevant snip of the class being serialized: class Person : ISerializable { private Dictionary<Relation,List<int>> Relationships = new Dictionary<Relation,List<int>>(); public Person(SerializationInfo info, StreamingContext context) { this.Relationships = (Dictionary<Relation, List<int>>) info.GetValue("Relationships", typeof(Dictionary<Relation, List<int>>)); } public void GetObjectData(SerializationInfo info, StreamingContext context) { info.AddValue("Relationships", this.Relationships); } } Note, this is binary serialization. Everything else in the project serializes and deserialzes correctly.

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  • Q on Python serialization/deserialization

    - by neil
    What chances do I have to instantiate, keep and serialize/deserialize to/from binary data Python classes reflecting this pattern (adopted from RFC 2246 [TLS]): enum { apple, orange } VariantTag; struct { uint16 number; opaque string<0..10>; /* variable length */ } V1; struct { uint32 number; opaque string[10]; /* fixed length */ } V2; struct { select (VariantTag) { /* value of selector is implicit */ case apple: V1; /* VariantBody, tag = apple */ case orange: V2; /* VariantBody, tag = orange */ } variant_body; /* optional label on variant */ } VariantRecord; Basically I would have to define a (variant) class VariantRecord, which varies depending on the value of VariantTag. That's not that difficult. The challenge is to find a most generic way to build a class, which serializes/deserializes to and from a byte stream... Pickle, Google protocol buffer, marshal is all not an option. I made little success with having an explicit "def serialize" in my class, but I'm not very happy with it, because it's not generic enough. I hope I could express the problem. My current solution in case VariantTag = apple would look like this, but I don't like it too much import binascii import struct class VariantRecord(object): def __init__(self, number, opaque): self.number = number self.opaque = opaque def serialize(self): out = struct.pack('>HB%ds' % len(self.opaque), self.number, len(self.opaque), self.opaque) return out v = VariantRecord(10, 'Hello') print binascii.hexlify(v.serialize()) >> 000a0548656c6c6f Regards

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  • Is a red-black tree my ideal data structure?

    - by Hugo van der Sanden
    I have a collection of items (big rationals) that I'll be processing. In each case, processing will consist of removing the smallest item in the collection, doing some work, and then adding 0-2 new items (which will always be larger than the removed item). The collection will be initialised with one item, and work will continue until it is empty. I'm not sure what size the collection is likely to reach, but I'd expect in the range 1M-100M items. I will not need to locate any item other than the smallest. I'm currently planning to use a red-black tree, possibly tweaked to keep a pointer to the smallest item. However I've never used one before, and I'm unsure whether my pattern of use fits its characteristics well. 1) Is there a danger the pattern of deletion from the left + random insertion will affect performance, eg by requiring a significantly higher number of rotations than random deletion would? Or will delete and insert operations still be O(log n) with this pattern of use? 2) Would some other data structure give me better performance, either because of the deletion pattern or taking advantage of the fact I only ever need to find the smallest item? Update: glad I asked, the binary heap is clearly a better solution for this case, and as promised turned out to be very easy to implement. Hugo

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  • Returning recursive ternary freaks out

    - by David Titarenco
    Hi, assume this following function: int binaryTree::findHeight(node *n) { if (n == NULL) { return 0; } else { return 1 + max(findHeight(n->left), findHeight(n->right)); } } Pretty standard recursive treeHeight function for a given binary search tree binaryTree. Now, I was helping a friend (he's taking an algorithms course), and I ran into some weird issue with this function that I couldn't 100% explain to him. With max being defined as max(a,b) ((a)>(b)?(a):(b)) (which happens to be the max definition in windef.h), the recursive function freaks out (it runs something like n^n times where n is the tree height). This obviously makes checking the height of a tree with 3000 elements take very, very long. However, if max is defined via templating, like std does it, everything is okay. So using std::max fixed his problem. I just want to know why. Also, why does the countLeaves function work fine, using the same programmatic recursion? int binaryTree::countLeaves(node *n) { if (n == NULL) { return 0; } else if (n->left == NULL && n->right == NULL) { return 1; } else { return countLeaves(n->left) + countLeaves(n->right); } } Is it because in returning the ternary function, the values a => countLeaves(n->left) and b => countLeaves(n->right) were recursively double called simply because they were the resultants? Thank you!

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  • Why can't the 'NonSerialized' attribute be used at the class level? How to prevent serialization of

    - by ck
    I have a data object that is deep-cloned using a binary serialization. This data object supports property changed events, for example, PriceChanged. Let's say I attached a handler to PriceChanged. When the code attempts to serialize PriceChanged, it throws an exception that the handler isn't marked as serializable. My alternatives: I can't easily remove all handlers from the event before serialization I don't want to mark the handler as serializable because I'd have to recursively mark all the handlers dependencies as well. I don't want to mark PriceChanged as NonSerialized - there are tens of events like this that could potentially have handlers. Ideally, I'd like .NET to just stop going down the object graph at that point and make that a 'leaf'. So why can't I just mark the handler class as 'NonSerialized'? -- I finally worked around this problem by making the handler implement ISerializable and doing nothing in the serialize constructor/ GetDataObject method. But, the handler still is serialized, just with all its dependencies set to null - so I had to account for that as well. Is there a better way to prevent serialization of an entire class?

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  • Log 2 N generic comparison tree

    - by Morano88
    Hey! I'm working on an algorithm for Redundant Binary Representation (RBR) where every two bits represent a digit. I designed the comparator that takes 4 bits and gives out 2 bits. I want to make the comparison in log 2 n so If I have X and Y .. I compare every 2 bits of X with every 2 bits of Y. This is smooth if the number of bits of X or Y equals n where (n = 2^X) i.e n = 2,4,8,16,32,... etc. Like this : However, If my input let us say is 6 or 10 .. then it becomes not smooth and I have to take into account some odd situations like this : I have a shallow experience in algorithms .. is there a generic way to do it .. so at the end I get only 2 bits no matter what the input is ? I just need hints or pseudo-code. If my question is not appropriate here .. so feel free to flag it or tell me to remove it. I'm using VHDL by the way!

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  • Count bits used in int

    - by sigvardsen
    If you have the binary number 10110 how can I get it to return 5? e.g a number that tells how many bits are used? There are some likewise examples listed below: 101 should return 3 000000011 should return 2 11100 should return 5 101010101 should return 9 How can this be obtained the easiest way in Java? I have come up with the following method but can i be done faster: public static int getBitLength(int value) { int l = 1; if (value >> 16 > 0) { value = value >> 16; l += 16; } if (value >> 8 > 0) { value = value >> 8; l += 8; } if (value >> 4 > 0) { value = value >> 4; l += 4; } if (value >> 2 > 0) { value = value >> 2; l += 2; } if (value >> 1 > 0) { value = value >> 1; l += 1; } return l; }

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  • problem with binarysearch algorithm

    - by arash
    hi friends,the code below belongs to binary search algorithm,user enter numbers in textbox1 and enter the number that he want to fing with binarysearch in textbox2.i have a problem with it,that is when i enter for example 15,21 in textbox1 and enter 15 in textbox2 and put brakpoint on the line i commented below,and i understood that it doesnt put the number in textbox2 in searchnums(commented),for more explanation i comment in code.thanks in advance public void button1_Click(object sender, EventArgs e) { int searchnums = Convert.ToInt32(textBox2.Text);//the problem is here,the value in textbox2 doesnt exist in searchnums and it has value 0. int result = binarysearch(searchnums); MessageBox.Show(result.ToString()); } public int binarysearch(int searchnum) { string[] source = textBox1.Text.Split(','); int[] nums = new int[source.Length]; for (int i = 0; i < source.Length; i++) { nums[i] = Convert.ToInt32(source[i]); } int first =0; int last = nums.Length; int mid = (int)Math.Floor(nums.Length / 2.0); while (1<= nums.Length) { if (searchnum < nums[mid]) { last = mid - 1; } if (searchnum > nums[mid]) { first = mid + 1; } else { return nums[mid]; } } return -1; }

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  • Detection of negative integers using bit operations

    - by Nawaz
    One approach to check if a given integer is negative or not, could be this: (using bit operations) int num_bits = sizeof(int) * 8; //assuming 8 bits per byte! int sign_bit = given_int & (1 << (num_bits-1)); //sign_bit is either 1 or 0 if ( sign_bit ) { cout << "given integer is negative"<<endl; } else { cout << "given integer is positive"<<endl; } The problem with this solution is that number of bits per byte couldn't be 8, it could be 9,10, 11 even 16 or 40 bits per byte. Byte doesn't necessarily mean 8 bits! Anyway, this problem can be easily fixed by writing, //CHAR_BIT is defined in limits.h int num_bits = sizeof(int) * CHAR_BIT; //no assumption. It seems fine now. But is it really? Is this Standard conformant? What if the negative integer is not represented as 2's complement? What if it's representation in a binary numeration system that doesn't necessitate only negative integers to have 1 in it's most significant bit? Can we write such code that will be both portable and standard conformant? Related topics: Size of Primitive data types Why is a boolean 1 byte and not 1 bit of size?

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  • Extract history from Korn shell

    - by Luc
    I am not happy about the history file in binary format of the Korn shell. I like to "collect" some of my command lines, many of them actually, and for a long time. I'm talking about years. That doesn't seem easy in Korn because the history file is not plain text so I can't edit it, and a lot of junk is piling up in it. By "junk" I mean lines that I don'twant to keep, like 'cat' or 'man'. So I added these lines to my .profile: fc -ln 1 9999 ~/khistory.txt source ~/loghistory.sh ~/khistory.txt loghistory.sh contains a handful of sed and sort commands that gets rid of a lot of the junk. But apparently it is forbidden to run fc in the .profile file. I can't login whenever I do, the shell exits right away with signal 11. So I removed that 'fc -l' line from my .profile file and added it to the loghistory.sh script, but the shell still crashes. I also tried this line in my .profile: strings ~/.sh_history ~/khistory.txt source ~/loghistory.sh That doesn't crash, but the output is printed with an additional, random character in the beginning of many lines. I can run 'fc -l' on the command line, but that's no good. I need to automate that. But how? How can I extract my ksh history as plain text? TIA

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  • Algorithm for assigning a unique series of bits for each user?

    - by Mark
    The problem seems simple at first: just assign an id and represent that in binary. The issue arises because the user is capable of changing as many 0 bits to a 1 bit. To clarify, the hash could go from 0011 to 0111 or 1111 but never 1010. Each bit has an equal chance of being changed and is independent of other changes. What would you have to store in order to go from hash - user assuming a low percentage of bit tampering by the user? I also assume failure in some cases so the correct solution should have an acceptable error rate. I would an estimate the maximum number of bits tampered with would be about 30% of the total set. I guess the acceptable error rate would depend on the number of hashes needed and the number of bits being set per hash. I'm worried with enough manipulation the id can not be reconstructed from the hash. The question I am asking I guess is what safe guards or unique positioning systems can I use to ensure this happens.

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  • In-order tree traversal

    - by Chris S
    I have the following text from an academic course I took a while ago about in-order traversal (they also call it pancaking) of a binary tree (not BST): In-order tree traversal Draw a line around the outside of the tree. Start to the left of the root, and go around the outside of the tree, to end up to the right of the root. Stay as close to the tree as possible, but do not cross the tree. (Think of the tree — its branches and nodes — as a solid barrier.) The order of the nodes is the order in which this line passes underneath them. If you are unsure as to when you go “underneath” a node, remember that a node “to the left” always comes first. Here's the example used (slightly different tree from below) However when I do a search on google, I get a conflicting definition. For example the wikipedia example: Inorder traversal sequence: A, B, C, D, E, F, G, H, I (leftchild,rootnode,right node) But according to (my understanding of) definition #1, this should be A, B, D, C, E, F, G, I, H Can anyone clarify which definition is correct? They might be both describing different traversal methods, but happen to be using the same name. I'm having trouble believing the peer-reviewed academic text is wrong, but can't be certain.

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  • Tree iterator, can you optimize this any further?

    - by Ron
    As a follow up to my original question about a small piece of this code I decided to ask a follow up to see if you can do better then what we came up with so far. The code below iterates over a binary tree (left/right = child/next ). I do believe there is room for one less conditional in here (the down boolean). The fastest answer wins! The cnt statement can be multiple statements so lets make sure this appears only once The child() and next() member functions are about 30x as slow as the hasChild() and hasNext() operations. Keep it iterative <-- dropped this requirement as the recursive solution presented was faster. This is C++ code visit order of the nodes must stay as they are in the example below. ( hit parents first then the children then the 'next' nodes). BaseNodePtr is a boost::shared_ptr as thus assignments are slow, avoid any temporary BaseNodePtr variables. Currently this code takes 5897ms to visit 62200000 nodes in a test tree, calling this function 200,000 times. void processTree (BaseNodePtr current, unsigned int & cnt ) { bool down = true; while ( true ) { if ( down ) { while (true) { cnt++; // this can/will be multiple statesments if (!current->hasChild()) break; current = current->child(); } } if ( current->hasNext() ) { down = true; current = current->next(); } else { down = false; current = current->parent(); if (!current) return; // done. } } }

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  • What is wrong with this code for reading binary files? [on hold]

    - by qed
    What is wrong with this code for reading binary files? It compiles OK, but will not print out the file as planned, in fact, it prints nothing at all. #include <iostream> #include <fstream> int main(int argc, const char *argv[]) { if (argc < 2) { ::std::cerr << "usage: " << argv[0] << " <filename>\n"; return 1; } ::std::basic_ifstream<unsigned char> in(argv[1], ::std::ios::binary); unsigned char uc; while (in.get(uc)) { printf("%02X ", uc); } // TODO: error handling, in case the file could not be opened or read return 0; }

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  • Designing a database file format

    - by RoliSoft
    I would like to design my own database engine for educational purposes, for the time being. Designing a binary file format is not hard nor the question, I've done it in the past, but while designing a database file format, I have come across a very important question: How to handle the deletion of an item? So far, I've thought of the following two options: Each item will have a "deleted" bit which is set to 1 upon deletion. Pro: relatively fast. Con: potentially sensitive data will remain in the file. 0x00 out the whole item upon deletion. Pro: potentially sensitive data will be removed from the file. Con: relatively slow. Recreating the whole database. Pro: no empty blocks which makes the follow-up question void. Con: it's a really good idea to overwrite the whole 4 GB database file because a user corrected a typo. I will sell this method to Twitter ASAP! Now let's say you already have a few empty blocks in your database (deleted items). The follow-up question is how to handle the insertion of a new item? Append the item to the end of the file. Pro: fastest possible. Con: file will get huge because of all the empty blocks that remain because deleted items aren't actually deleted. Search for an empty block exactly the size of the one you're inserting. Pro: may get rid of some blocks. Con: you may end up scanning the whole file at each insert only to find out it's very unlikely to come across a perfectly fitting empty block. Find the first empty block which is equal or larger than the item you're inserting. Pro: you probably won't end up scanning the whole file, as you will find an empty block somewhere mid-way; this will keep the file size relatively low. Con: there will still be lots of leftover 0x00 bytes at the end of items which were inserted into bigger empty blocks than they are. Rigth now, I think the first deletion method and the last insertion method are probably the "best" mix, but they would still have their own small issues. Alternatively, the first insertion method and scheduled full database recreation. (Probably not a good idea when working with really large databases. Also, each small update in that method will clone the whole item to the end of the file, thus accelerating file growth at a potentially insane rate.) Unless there is a way of deleting/inserting blocks from/to the middle of the file in a file-system approved way, what's the best way to do this? More importantly, how do databases currently used in production usually handle this?

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  • What kind of string is this? What can I do in php to read it?

    - by kevin
    This is a string (see below, after the dashed line) in a database.inf file for a free program I downloaded that lists some websites. The file is plain text as you can see , but there is a string after it that looks base64 encoded (due to the end chars of ==). But b64_decoding it gives giberish. I wanted to decode it so I could add to the list of sites it had (the program lists a bunch of sites and data about them which I can read in the GUI) and to do that I need to decode this, add to it, and re-encode it. I think the program uses .net since I think the .net library was required on install, but I know nothing of the original source language. I am using php to figure out if there is a simple way to read this. I have tried using unpack, binhex, base_convert, etc as I suspect the file is binary at some level, but I am lost. Nothing illegal, just wanting to know what it is and if I can add a few things to it to make it more useful for me. here is the file - any ideas how to decode and recode this for playing with? Site List file size: 62139 db version: 13 generated: 2010-04-27 11:53:40 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  • C#: My World Clock

    - by Bruce Eitman
    [Placeholder:  I will post the entire project soon] I have been working on cleaning my office of 8 years of stuff from several engineers working on many projects.  It turns out that we have a few extra single board computers with displays, so at the end of the day last Friday I though why not create a little application to display the time, you know, a clock.  How difficult could that be?  It turns out that it is quite simple – until I decided to gold plate the project by adding time displays for our offices around the world. I decided to use C#, which actually made creating the main clock quite easy.   The application was simply a text box and a timer.  I set the timer to fire a couple of times a second, and when it does use a DateTime object to get the current time and retrieve a string to display. And I could have been done, but of course that gold plating came up.   Seems simple enough, simply offset the time from the local time to the location that I want the time for and display it.    Sure enough, I had the time displayed for UK, Italy, Kansas City, Japan and China in no time at all. But it is October, and for those of us still stuck with Daylight Savings Time, we know that the clocks are about to change.   My first attempt was to simply check to see if the local time was DST or Standard time, then change the offset for China.  China doesn’t have Daylight Savings Time. If you know anything about the time changes around the world, you already know that my plan is flawed – in a big way.   It turns out that the transitions in and out of DST take place at different times around the world.   If you didn’t know that, do a quick search for “Daylight Savings” and you will find many WEB sites dedicated to tracking the time changes dates, and times. Now the real challenge of this application; how do I programmatically find out when the time changes occur and handle them correctly?  After a considerable amount of research it turns out that the solution is to read the data from the registry and parse it to figure out when the time changes occur. Reading Time Change Information from the Registry Reading the data from the registry is simple, using the data is a little more complicated.  First, reading from the registry can be done like:             byte[] binarydata = (byte[])Registry.GetValue("HKEY_LOCAL_MACHINE\\Time Zones\\Eastern Standard Time", "TZI", null);   Where I have hardcoded the registry key for example purposes, but in the end I will use some variables.   We now have a binary blob with the data, but it needs to be converted to use the real data.   To start we will need a couple of structs to hold the data and make it usable.   We will need a SYSTEMTIME and REG_TZI_FORMAT.   You may have expected that we would need a TIME_ZONE_INFORMATION struct, but we don’t.   The data is stored in the registry as a REG_TZI_FORMAT, which excludes some of the values found in TIME_ZONE_INFORMATION.     struct SYSTEMTIME     {         internal short wYear;         internal short wMonth;         internal short wDayOfWeek;         internal short wDay;         internal short wHour;         internal short wMinute;         internal short wSecond;         internal short wMilliseconds;     }       struct REG_TZI_FORMAT     {         internal long Bias;         internal long StdBias;         internal long DSTBias;         internal SYSTEMTIME StandardStart;         internal SYSTEMTIME DSTStart;     }   Now we need to convert the binary blob to a REG_TZI_FORMAT.   To do that I created the following helper functions:         private void BinaryToSystemTime(ref SYSTEMTIME ST, byte[] binary, int offset)         {             ST.wYear = (short)(binary[offset + 0] + (binary[offset + 1] << 8));             ST.wMonth = (short)(binary[offset + 2] + (binary[offset + 3] << 8));             ST.wDayOfWeek = (short)(binary[offset + 4] + (binary[offset + 5] << 8));             ST.wDay = (short)(binary[offset + 6] + (binary[offset + 7] << 8));             ST.wHour = (short)(binary[offset + 8] + (binary[offset + 9] << 8));             ST.wMinute = (short)(binary[offset + 10] + (binary[offset + 11] << 8));             ST.wSecond = (short)(binary[offset + 12] + (binary[offset + 13] << 8));             ST.wMilliseconds = (short)(binary[offset + 14] + (binary[offset + 15] << 8));         }             private REG_TZI_FORMAT ConvertFromBinary(byte[] binarydata)         {             REG_TZI_FORMAT RTZ = new REG_TZI_FORMAT();               RTZ.Bias = binarydata[0] + (binarydata[1] << 8) + (binarydata[2] << 16) + (binarydata[3] << 24);             RTZ.StdBias = binarydata[4] + (binarydata[5] << 8) + (binarydata[6] << 16) + (binarydata[7] << 24);             RTZ.DSTBias = binarydata[8] + (binarydata[9] << 8) + (binarydata[10] << 16) + (binarydata[11] << 24);             BinaryToSystemTime(ref RTZ.StandardStart, binarydata, 4 + 4 + 4);             BinaryToSystemTime(ref RTZ.DSTStart, binarydata, 4 + 16 + 4 + 4);               return RTZ;         }   I am the first to admit that there may be a better way to get the settings from the registry and into the REG_TXI_FORMAT, but I am not a great C# programmer which I have said before on this blog.   So sometimes I chose brute force over elegant. Now that we have the Bias information and the start date information, we can start to make sense of it.   The bias is an offset, in minutes, from local time (if already in local time for the time zone in question) to get to UTC – or as Microsoft defines it: UTC = local time + bias.  Standard bias is an offset to adjust for standard time, which I think is usually zero.   And DST bias is and offset to adjust for daylight savings time. Since we don’t have the local time for a time zone other than the one that the computer is set to, what we first need to do is convert local time to UTC, which is simple enough using:                 DateTime.Now.ToUniversalTime(); Then, since we have UTC we need to do a little math to alter the formula to: local time = UTC – bias.  In other words, we need to subtract the bias minutes. I am ahead of myself though, the standard and DST start dates really aren’t dates.   Instead they indicate the month, day of week and week number of the time change.   The dDay member of SYSTEM time will be set to the week number of the date change indicating that the change happens on the first, second… day of week of the month.  So we need to convert them to dates so that we can determine which bias to use, and when to change to a different bias.   To do that, I wrote the following function:         private DateTime SystemTimeToDateTimeStart(SYSTEMTIME Time, int Year)         {             DayOfWeek[] Days = { DayOfWeek.Sunday, DayOfWeek.Monday, DayOfWeek.Tuesday, DayOfWeek.Wednesday, DayOfWeek.Thursday, DayOfWeek.Friday, DayOfWeek.Saturday };             DateTime InfoTime = new DateTime(Year, Time.wMonth, Time.wDay == 1 ? 1 : ((Time.wDay - 1) * 7) + 1, Time.wHour, Time.wMinute, Time.wSecond, DateTimeKind.Utc);             DateTime BestGuess = InfoTime;             while (BestGuess.DayOfWeek != Days[Time.wDayOfWeek])             {                 BestGuess = BestGuess.AddDays(1);             }             return BestGuess;         }   SystemTimeToDateTimeStart gets two parameters; a SYSTEMTIME and a year.   The reason is that we will try this year and next year because we are interested in start dates that are in the future, not the past.  The function starts by getting a new Datetime with the first possible date and then looking for the correct date. Using the start dates, we can then determine the correct bias to use, and the next date that time will change:             NextTimeChange = StandardChange;             CurrentBias = TimezoneSettings.Bias + TimezoneSettings.DSTBias;             if (DSTChange.Year != 1 && StandardChange.Year != 1)             {                 if (DSTChange.CompareTo(StandardChange) < 0)                 {                     NextTimeChange = DSTChange;                     CurrentBias = TimezoneSettings.StdBias + TimezoneSettings.Bias;                 }             }             else             {                 // I don't like this, but it turns out that China Standard Time                 // has a DSTBias of -60 on every Windows system that I tested.                 // So, if no DST transitions, then just use the Bias without                 // any offset                 CurrentBias = TimezoneSettings.Bias;             }   Note that some time zones do not change time, in which case the years will remain set to 1.   Further, I found that the registry settings are actually wrong in that the DST Bias is set to -60 for China even though there is not DST in China, so I ignore the standard and DST bias for those time zones. There is one thing that I have not solved, and don’t plan to solve.  If the time zone for this computer changes, this application will not update the clock using the new time zone.  I tell  you this because you may need to deal with it – I do not because I won’t let the user get to the control panel applet to change the timezone. Copyright © 2012 – Bruce Eitman All Rights Reserved

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  • How do I read binary C++ protobuf data using Python protobuf?

    - by nbolton
    The Python version of Google protobuf gives us only: SerializeAsString() Where as the C++ version gives us both: SerializeToArray(...) SerializeAsString() We're writing to our C++ file in binary format, and we'd like to keep it this way. That said, is there a way of reading the binary data into Python and parsing it as if it were a string? Is this the correct way of doing it? binary = get_binary_data() binary_size = get_binary_size() string = None for i in range(len(binary_size)): string += i message = new MyMessage() message.ParseFromString(string) Update: Here's a new example, and a problem: message_length = 512 file = open('foobars.bin', 'rb') eof = False while not eof: data = file.read(message_length) eof = not data if not eof: foo_bar = FooBar() foo_bar.ParseFromString(data) When we get to the foo_bar.ParseFromString(data) line, I get this error: Exception Type: DecodeError Exception Value: Too many bytes when decoding varint. Update 2: It turns out, that the padding on the binary data was throwing protobuf off; too many bytes were being sent in, as the message suggests (in this case it was referring to the padding). This padding comes from using the C++ protobuf function, SerializeToArray on a fixed-length buffer. To eliminate this, I have used this temproary code: message_length = 512 file = open('foobars.bin', 'rb') eof = False while not eof: data = file.read(message_length) eof = not data string = '' for i in range(0, len(data)): byte = data[i] if byte != '\xcc': # yuck! string += data[i] if not eof: foo_bar = FooBar() foo_bar.ParseFromString(string) There is a design flaw here I think. I will re-implement my C++ code so that it writes variable length arrays to the binary file. As advised by the protobuf documentation, I will prefix each message with it's binary size so that I know how much to read when I'm opening the file with Python.

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  • two's complement, why the name "two"

    - by lenatis
    i know unsigned,two's complement, ones' complement and sign magnitude, and the difference between these, but what i'm curious about is: why it's called two's(or ones') complement, so is there a more generalize N's complement? in which way did these genius deduce such a natural way to represent negative numbers?

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  • Which platform can we expect one's complement being used there?

    - by Jian Lin
    For some questions such as checking whether a number is odd or even, I noted the comment, a & 1 won't work when it is a one's complement machine or when the code is ported to a platform that uses one's complement. Since 30 years ago on the Superboard, TRS-80, Apple II, I haven't seen a system with one's complement. Are there popular systems that use one's complement still, or do we have some cell phone or mobile device that uses one's complement?

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