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  • Can a binary tree or tree be always represented in a Database as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • Can a binary tree or tree be always represented in a Database table as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person under him can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • Implementation of a distance matrix of a binary tree that is given in the adjacency-list representation

    - by Denise Giubilei
    Given this problem I need an O(n²) implementation of this algorithm: "Let v be an arbitrary leaf in a binary tree T, and w be the only vertex connected to v. If we remove v, we are left with a smaller tree T'. The distance between v and any vertex u in T' is 1 plus the distance between w and u." This is a problem and solution of a Manber's exercise (Exercise 5.12 from U. Manber, Algorithms: A Creative Approach, Addison-Wesley (1989).). The thing is that I can't deal properly with the adjacency-list representation so that the implementation of this algorithm results in a real O(n²) complexity. Any ideas of how the implementation should be? Thanks.

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  • Any efficient way to read datas from large binary file?

    - by limi
    Hi, I need to handle tens of Gigabytes data in one binary file. Each record in the data file is variable length. So the file is like: <len1><data1><len2><data2>..........<lenN><dataN> The data contains integer, pointer, double value and so on. I found python can not even handle this situation. There is no problem if I read the whole file in memory. It's fast. But it seems the struct package is not good at performance. It almost stuck on unpack the bytes. Any help is appreciated. Thanks.

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  • Is there something special about the number 65535?

    - by Nick Rosencrantz
    2¹6-1 & 25 = 25 (or? obviously ?) A developer asked me today what is bitwise 65535 & 32 i.e. 2¹6-1 & 25 = ? I thought at first spontaneously 32 but it seemed to easy whereupon I thought for several minutes and then answered 32. 32 seems to have been the correct answer but how? 65535=2¹6-1=1111111111111111 (but it doesn't seem right since this binary number all ones should be -1(?)), 32 = 100000 but I could not convert that in my head whereupon I anyway answered 32 since I had to answer something. Is the answer 32 in fact trivial? Is in the same way 2¹6-1 & 25-1 =31? Why did the developer ask me about exactly 65535? Binary what I was asked to evaluate was 1111111111111111 & 100000 but I don't understand why 1111111111111111 is not -1. Shouldn't it be -1? Is 65535 a number that gives overflow and how do I know that?

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  • What would be the time complexity of counting the number of all structurally different binary trees?

    - by ktslwy
    Using the method presented here: http://cslibrary.stanford.edu/110/BinaryTrees.html#java 12. countTrees() Solution (Java) /** For the key values 1...numKeys, how many structurally unique binary search trees are possible that store those keys? Strategy: consider that each value could be the root. Recursively find the size of the left and right subtrees. */ public static int countTrees(int numKeys) { if (numKeys <=1) { return(1); } else { // there will be one value at the root, with whatever remains // on the left and right each forming their own subtrees. // Iterate through all the values that could be the root... int sum = 0; int left, right, root; for (root=1; root<=numKeys; root++) { left = countTrees(root-1); right = countTrees(numKeys - root); // number of possible trees with this root == left*right sum += left*right; } return(sum); } } I have a sense that it might be n(n-1)(n-2)...1, i.e. n!

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  • Why is this attempt at a binary search crashing?

    - by Ian Campbell
    I am fairly new to the concept of a binary search, and am trying to write a program that does this in Java for personal practice. I understand the concept of this well, but my code is not working. There is a run-time exception happening in my code that just caused Eclipse, and then my computer, to crash... there are no compile-time errors here though. Here is what I have so far: public class BinarySearch { // instance variables int[] arr; int iterations; // constructor public BinarySearch(int[] arr) { this.arr = arr; iterations = 0; } // instance method public int findTarget(int targ, int[] sorted) { int firstIndex = 1; int lastIndex = sorted.length; int middleIndex = (firstIndex + lastIndex) / 2; int result = sorted[middleIndex - 1]; while(result != targ) { if(result > targ) { firstIndex = middleIndex + 1; middleIndex = (firstIndex + lastIndex) / 2; result = sorted[middleIndex - 1]; iterations++; } else { lastIndex = middleIndex + 1; middleIndex = (firstIndex + lastIndex) / 2; result = sorted[middleIndex - 1]; iterations++; } } return result; } // main method public static void main(String[] args) { int[] sortedArr = new int[] { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 }; BinarySearch obj = new BinarySearch(sortedArr); int target = sortedArr[8]; int result = obj.findTarget(target, sortedArr); System.out.println("The original target was -- " + target + ".\n" + "The result found was -- " + result + ".\n" + "This took " + obj.iterations + " iterations to find."); } // end of main method } // end of class BinarySearch

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  • adding virtual function to the end of the class declaration avoids binary incompatibility?

    - by bob
    Could someone explain to me why adding a virtual function to the end of a class declaration avoids binary incompatibility? If I have: class A { public: virtual ~A(); virtual void someFuncA() = 0; virtual void someFuncB() = 0; virtual void other1() = 0; private: int someVal; }; And later modify this function to: class A { public: virtual ~A(); virtual void someFuncA(); virtual void someFuncB(); virtual void someFuncC(); virtual void other1() = 0; private: int someVal; }; I get a coredump from another .so compiled against the previous declaration. But if I put someFuncC() at the end of the class declaration (after "int someVal"), I don't see coredump anymore. Could someone tell me why this is? And does this trick always work? PS. compiler is gcc, does this work with other compilers?

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  • How to get the size of a binary tree ?

    - by Andrei Ciobanu
    I have a very simple binary tree structure, something like: struct nmbintree_s { unsigned int size; int (*cmp)(const void *e1, const void *e2); void (*destructor)(void *data); nmbintree_node *root; }; struct nmbintree_node_s { void *data; struct nmbintree_node_s *right; struct nmbintree_node_s *left; }; Sometimes i need to extract a 'tree' from another and i need to get the size to the 'extracted tree' in order to update the size of the initial 'tree' . I was thinking on two approaches: 1) Using a recursive function, something like: unsigned int nmbintree_size(struct nmbintree_node* node) { if (node==NULL) { return(0); } return( nmbintree_size(node->left) + nmbintree_size(node->right) + 1 ); } 2) A preorder / inorder / postorder traversal done in an iterative way (using stack / queue) + counting the nodes. What approach do you think is more 'memory failure proof' / performant ? Any other suggestions / tips ? NOTE: I am probably going to use this implementation in the future for small projects of mine. So I don't want to unexpectedly fail :).

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  • Save and Load Dictionary from encrypted or unreadable or binary format ?

    - by Prix
    I have a dictionary like the follow: public Dictionary<int, SpawnList> spawnEntities = new Dictionary<int, SpawnList>(); The class being used is as follow: public class SpawnList { public int NpcID { get; set; } public string Name { get; set; } public int Level { get; set; } public int TitleID { get; set; } public int StaticID { get; set; } public entityType Status { get; set; } public int TypeA { get; set; } public int TypeB { get; set; } public int TypeC { get; set; } public int ZoneID { get; set; } public int Heading { get; set; } public float PosX { get; set; } public float PosY { get; set; } public float PosZ { get; set; } } /// <summary>Entity type enum.</summary> public enum entityType { Ally, Enemy, SummonPet, NPC, Object, Monster, Gatherable, Unknown } How could I save this Dictionary to either a binary or encrypted format so I could later Load it again into my application ? My limitation is .Net 3.5 can't use anything higher.

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  • How do I add an object to a binary tree based on the value of a member variable?

    - by Max
    How can I get a specific value from an object? I'm trying to get a value of an instance for eg. ListOfPpl newListOfPpl = new ListOfPpl(id, name, age); Object item = newListOfPpl; How can I get a value of name from an Object item?? Even if it is easy or does not interest you can anyone help me?? Edited: I was trying to build a binary tree contains the node of ListOfPpl, and need to sort it in the lexicographic. Here's my code for insertion on the node. Any clue?? public void insert(Object item){ Node current = root; Node follow = null; if(!isEmpty()){ root = new Node(item, null, null); return; }boolean left = false, right = false; while(current != null){ follow = current; left = false; right = false; //I need to compare and sort it if(item.compareTo(current.getFighter()) < 0){ current = current.getLeft(); left = true; }else { current = current.getRight(); right = true; } }if(left) follow.setLeft(new Node(item, null, null)); else follow.setRight(new Node(item, null, null)); }

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  • Mercurial hook to disallow committing large binary files

    - by hekevintran
    I want to have a Mercurial hook that will run before committing a transaction that will abort the transaction if a binary file being committed is greater than 1 megabyte. I found the following code which works fine except for one problem. If my changeset involves removing a file, this hook will throw an exception. The hook (I'm using pretxncommit = python:checksize.newbinsize): from mercurial import context, util from mercurial.i18n import _ import mercurial.node as dpynode '''hooks to forbid adding binary file over a given size Ensure the PYTHONPATH is pointing where hg_checksize.py is and setup your repo .hg/hgrc like this: [hooks] pretxncommit = python:checksize.newbinsize pretxnchangegroup = python:checksize.newbinsize preoutgoing = python:checksize.nopull [limits] maxnewbinsize = 10240 ''' def newbinsize(ui, repo, node=None, **kwargs): '''forbid to add binary files over a given size''' forbid = False # default limit is 10 MB limit = int(ui.config('limits', 'maxnewbinsize', 10000000)) tip = context.changectx(repo, 'tip').rev() ctx = context.changectx(repo, node) for rev in range(ctx.rev(), tip+1): ctx = context.changectx(repo, rev) print ctx.files() for f in ctx.files(): fctx = ctx.filectx(f) filecontent = fctx.data() # check only for new files if not fctx.parents(): if len(filecontent) > limit and util.binary(filecontent): msg = 'new binary file %s of %s is too large: %ld > %ld\n' hname = dpynode.short(ctx.node()) ui.write(_(msg) % (f, hname, len(filecontent), limit)) forbid = True return forbid The exception: $ hg commit -m 'commit message' error: pretxncommit hook raised an exception: apps/helpers/templatetags/include_extends.py@bced6272d8f4: not found in manifest transaction abort! rollback completed abort: apps/helpers/templatetags/include_extends.py@bced6272d8f4: not found in manifest! I'm not familiar with writing Mercurial hooks, so I'm pretty confused about what's going on. Why does the hook care that a file was removed if hg already knows about it? Is there a way to fix this hook so that it works all the time? Update (solved): I modified the hook to filter out files that were removed in the changeset. def newbinsize(ui, repo, node=None, **kwargs): '''forbid to add binary files over a given size''' forbid = False # default limit is 10 MB limit = int(ui.config('limits', 'maxnewbinsize', 10000000)) ctx = repo[node] for rev in xrange(ctx.rev(), len(repo)): ctx = context.changectx(repo, rev) # do not check the size of files that have been removed # files that have been removed do not have filecontexts # to test for whether a file was removed, test for the existence of a filecontext filecontexts = list(ctx) def file_was_removed(f): """Returns True if the file was removed""" if f not in filecontexts: return True else: return False for f in itertools.ifilterfalse(file_was_removed, ctx.files()): fctx = ctx.filectx(f) filecontent = fctx.data() # check only for new files if not fctx.parents(): if len(filecontent) > limit and util.binary(filecontent): msg = 'new binary file %s of %s is too large: %ld > %ld\n' hname = dpynode.short(ctx.node()) ui.write(_(msg) % (f, hname, len(filecontent), limit)) forbid = True return forbid

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  • how to create binary of iphone application

    - by CodeWriter
    hello, On reading the iTunesConnect_DeveloperGuide.pdf, i came across the term binary. This is the paragraph found in the pdf: To submit your application through iTunes Connect and get it posted on the App Store successfully, make sure you have the following: Application binary (includes 57px icon), large 512px icon for use on the App Store, primary screenshot, contract information, export compliance information and applica- tion metadata It is also mentioned that the binary should be a zipped file. Can anyone please explain what is a binary and how to create one? Thanks

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  • Moving Binary logs in Mysql to a different harddisk

    - by Darini
    This question is about Mysql Binary logging.We need to move the Binary logging to a different hard disk . What is the configuration change required in Mysql?.Currently Binary logs go into the same folder as the ibdata and there is a replication slave running which needs the binary logs

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  • IIS7 Binary Stream Error

    - by WDuffy
    I'm struggling to understand the problem I'm seeing so please accept my apology if the question is vague. I'm running a classic asp app in IIS7 and everything seems to work fine except for one issue that has me stumped. Basically, files can be downloaded from the server which is done using the sendBinary method of Persits ASP Upload component. This component works fine for uploading etc it's just the downloading I have a problem. The strange thing is, I cannot have a pure asp page that serves the binary file. Everything works fine in II6, but in II7 there is a strange problem. For example this does not work. <% SET objUpload = server.createObject("Persits.Upload") objUpload.sendBinary "D:\sites\file.pdf", true, "application/octet-binary", true SET objUpload = NOTHING %> However, if I put anything in front of the asp code the file is served fine. serve this <% SET objUpload = server.createObject("Persits.Upload") objUpload.sendBinary "D:\sites\file.pdf", true, "application/octet-binary", true SET objUpload = NOTHING %> I can also write something to the response stream first and it works fine. <% Response.Write("server this") SET objUpload = server.createObject("Persits.Upload") objUpload.sendBinary "D:\sites\file.pdf", true, "application/octet-binary", true SET objUpload = NOTHING %> Does anyone have any ideas of what could be causing this or has anyone ran into a similar situation? I'm sure it has something to do with the setup in IIS 7.

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  • Compiling vs using pre-built binaries performance?

    - by Nick Rosencrantz
    Will performance be better (quicker) if I manually compile the source for a software component for the actual machine that it will be used on, compared to if the source was compiled on another platform perhaps for many different architectures? I got some good results compiling source that I downloaded and I wonder whether this was due to compiling it instead of downloading a pre-compiled binary which is often the case with software updates.

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  • Tournament bracket method to put distance between teammates

    - by Fred Thomsen
    I am using a proper binary tree to simulate a tournament bracket. It's preferred any competitors in the bracket that are teammates don't meet each other until the later rounds. What is an efficient method in which I can ensure that teammates in the bracket have as much distance as possible from each other? Are there any other data structures besides a tree that would be better for this purpose? EDIT: There can be more than 2 teams represented in a bracket.

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  • How portable are Binaries compiled in Ubuntu?

    - by hiobs
    The title says it all, actually. But allow me to specify the question: Assuming I were to compile an application that uses libffi, libGL, dlfcn, and SDL, would said binary run on other Linux distributions with same architecture, etc? The reason I ask is because of the directory /usr/lib/i386-linux-gnu - I might be wrong, but I assume this directory is something rather Ubuntu-specific, no? So, how portable are binaries compiled on Ubuntu really?

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  • How to find the insertion point in an array using binary search?

    - by ????
    The basic idea of binary search in an array is simple, but it might return an "approximate" index if the search fails to find the exact item. (we might sometimes get back an index for which the value is larger or smaller than the searched value). For looking for the exact insertion point, it seems that after we got the approximate location, we might need to "scan" to left or right for the exact insertion location, so that, say, in Ruby, we can do arr.insert(exact_index, value) I have the following solution, but the handling for the part when begin_index >= end_index is a bit messy. I wonder if a more elegant solution can be used? (this solution doesn't care to scan for multiple matches if an exact match is found, so the index returned for an exact match may point to any index that correspond to the value... but I think if they are all integers, we can always search for a - 1 after we know an exact match is found, to find the left boundary, or search for a + 1 for the right boundary.) My solution: DEBUGGING = true def binary_search_helper(arr, a, begin_index, end_index) middle_index = (begin_index + end_index) / 2 puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " + "begin_index = #{begin_index}, end_index = #{end_index}, " + "middle_index = #{middle_index}" if DEBUGGING if arr[middle_index] == a return middle_index elsif begin_index >= end_index index = [begin_index, end_index].min return index if a < arr[index] && index >= 0 #careful because -1 means end of array index = [begin_index, end_index].max return index if a < arr[index] && index >= 0 return index + 1 elsif a > arr[middle_index] return binary_search_helper(arr, a, middle_index + 1, end_index) else return binary_search_helper(arr, a, begin_index, middle_index - 1) end end # for [1,3,5,7,9], searching for 6 will return index for 7 for insertion # if exact match is found, then return that index def binary_search(arr, a) puts "\nSearching for #{a} in #{arr}" if DEBUGGING return 0 if arr.empty? result = binary_search_helper(arr, a, 0, arr.length - 1) puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING return result end arr = [1,3,5,7,9] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [] b = 60 arr.insert(binary_search(arr, b), b) p arr and result: Searching for 6 in [1, 3, 5, 7, 9] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9] Searching for 6 in [1, 3, 5, 7, 9, 11] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9, 11] Searching for 60 in [1, 3, 5, 7, 9] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4 the result is 5, the index for value nil [1, 3, 5, 7, 9, 60] Searching for 60 in [1, 3, 5, 7, 9, 11] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5 the result is 6, the index for value nil [1, 3, 5, 7, 9, 11, 60] Searching for -60 in [1, 3, 5, 7, 9] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9] Searching for -60 in [1, 3, 5, 7, 9, 11] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9, 11] Searching for -60 in [1] a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 0, the index for value 1 [-60, 1] Searching for 60 in [1] a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 1, the index for value nil [1, 60] Searching for 60 in [] [60]

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  • How to compile scheme into native binary files ?

    - by Joe
    I am very new to scheme. And now I am trying to compile some scheme code into binary file which will be loaded faster into interpreter. (The interpreter is a hybrid interpreter)Some one told me that I can compile the code into native binary file and then load it into interperter. And my question is: 1. What is the native binary file? 2. How can I compile the scheme code into a native binary file? 3. How can I load native bianry file into scheme interpreter? Thanks in advance. Joe Suggested that I want to compile below code into native binary file: (define test (lambda() (display "this is a test")) And then load the bianry file into interpreter and call the function "test".

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  • C++ : Lack of Standardization at the Binary Level

    - by Nawaz
    Why ISO/ANSI didn't standardize C++ at the binary level? There are many portability issues with C++, which is only because of lack of it's standardization at the binary level. Don Box writes, (quoting from his book Essential COM, chapter COM As A Better C++) C++ and Portability Once the decision is made to distribute a C++ class as a DLL, one is faced with one of the fundamental weaknesses of C++, that is, lack of standardization at the binary level. Although the ISO/ANSI C++ Draft Working Paper attempts to codify which programs will compile and what the semantic effects of running them will be, it makes no attempt to standardize the binary runtime model of C++. The first time this problem will become evident is when a client tries to link against the FastString DLL's import library from a C++ developement environment other than the one used to build the FastString DLL. Are there more benefits Or loss of this lack of binary standardization?

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  • String or binary data would be truncated.

    - by Derek Dieter
    This error message is relatively straight forward. The way it normally happens is when you are trying to insert data from a table that contains values that have larger data lengths than the table you are trying to insert into. An example of this would be trying to insert data from a permanent table, into [...]

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  • Building package from binary files - what's wrong with my control file

    - by Hannes de Jager
    I'm busy trying to build a .deb package from the binaries of my application (non open source) and I'm having trouble getting the correct info to display in the Ubuntu Software Centre (when you click on the .deb file). Please see screenshot below of control file and Software Centre View. It seems like the package name and the package description is swapped. I'm expecting the part in bold to read "attix5pro" and not "Cloud backup agent". Can someone show my my mistake or guide me?

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  • Packaging MATLAB (or, more generally, a large binary, proprietary piece of software)

    - by nfirvine
    I'm trying to package MATLAB for internal distribution, but this could apply to any piece of software with the same architecture. In fact, I'm packaging multiple releases of MATLAB to be installed concurrently. Key things Very large installation size (~4 GB) Composed of a core, and several plugins (toolboxes) Initially, I created a single "source" package (matlab2011b) that builds several .debs (mainly matlab2011b-core and matlab2011b-toolbox-* for each toolbox). The control file is just the standard all: dh $@ There is no Makefile; only copying files. I use a number of debian/*.install files to specify files to copy from a copy of an installation to /usr/lib/. The problem is, every time I build the thing (say, to make a correction to the core package), it recopies every file listed in the *.install file to e.g debian/$packagename/usr/ (the build phase), and then has to bundle that into a .deb file. It takes a long time, on the order of hours, and is doing a lot of extra work. So my questions are: Can you make dh_install do a hardlink copy (like cp -l) to save time? (AFAICT from the man page, no.) Maybe I should just get it to do this in the Makefile? (That's gonna b e big Makefile.) Can you make debuild only rebuild .debs that need rebuilding? Or specify which .debs to rebuild? Is my approach completely stupid? Should I break each of the toolboxes into its own source package too? (I'll have to do some silly templating or something, because there's hundreds of them. :/)

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