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  • A space-efficient guest filesystem for grow-as-needed virtual disks ?

    - by Steve Schnepp
    A common practice is to use non-preallocated virtual disks. Since they only grow as needed, it makes them perfect for fast backup, overallocation and creation speed. Since file systems are usually based on physical disks they have the tendency to use the whole area available1 in order to increase the speed2 or reliability3. I'm searching a filesystem that does the exact opposite : try to touch the minimum blocks need by an aggressive block reuse. I would happily trade some performance for space usage. There is already a similar question, but it is rather general. I have very specific goal : space-efficiency. 1. Like page caching uses all the free physical memory 2. Canonical example : online defragmentation 3. Canonical example : snapshotting

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  • How do you get more space on Live Mesh/Live Sync? [closed]

    - by Narcolapser
    I'm looking into corporate data backs for my company's dealers. each dealership will have data back up demands ranging from 100mb to 20gb. We are an entirely Microsoft solution so when I was asked to look into back ups, of course I would look to Micro$oft. even if we have too buy this space, is there a way to get more space on Live Mesh/Live Sync (Live Mync hehe)? The 5 gb that Mesh provides or the 2gb that Sync provides isn't enough for our larger dealerships. The 25gb that SkyDrive provides is probably enough for now, but I don't know if it will be in the future. However, SkyDrive is not automatically synced. So it isn't a viable option anyway.

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  • Why some recovery tools are still able to find deleted files after I purge Recycle Bin, defrag the disk and zero-fill free space?

    - by Ivan
    As far as I understand, when I delete (without using Recycle Bin) a file, its record is removed from the file system table of contents (FAT/MFT/etc...) but the values of the disk sectors which were occupied by the file remain intact until these sectors are reused to write something else. When I use some sort of erased files recovery tool, it reads those sectors directly and tries to build up the original file. In this case, what I can't understand is why recovery tools are still able to find deleted files (with reduced chance of rebuilding them though) after I defragment the drive and overwrite all the free space with zeros. Can you explain this? I thought zero-overwritten deleted files can be only found by means of some special forensic lab magnetic scan hardware and those complex wiping algorithms (overwriting free space multiple times with random and non-random patterns) only make sense to prevent such a physical scan to succeed, but practically it seems that plain zero-fill is not enough to wipe all the tracks of deleted files. How can this be?

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  • A space-efficient filesystem for grow-as-needed virtual disks ?

    - by Steve Schnepp
    A common practice is to use non-preallocated virtual disks. Since they only grow as needed, it makes them perfect for fast backup, overallocation and creation speed. Since file systems are usually based on physical disks they have the tendency to use the whole area available1 in order to increase the speed2 or reliability3. I'm searching a filesystem that does the exact opposite : try to touch the minimum blocks need by an aggressive block reuse. I would happily trade some performance for space usage. There is already a similar question, but it is rather general. I have very specific goal : space-efficiency. 1. Like page caching uses all the free physical memory 2. Canonical example : online defragmentation 3. Canonical example : snapshotting

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  • Drive system file size

    - by rezx
    When i made a new drive it take some space for system file FAT32 take the less space, then NTFS, then ext4 my question how to know the space will be taken for the system before make the drive, if the drive 1giga or 100giga for FAT32, NTFS, ext4. Edit: when make 10MB drive with FAT32 the size shown 9.9 when make 10MB drive with ext4 the size shown 8.1 the same thing with the bigger size there always some space used and there is no files on the drive, so where this space go, if it for the filesystem how i can calculate the space that will be taken before format the drive

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  • What is stable as Ubuntu in kernel space, but also cutting-edge as Fedora in user land?

    - by HRJ
    I am a long time Fedora user (since Fedora 6). Previously I have used Gentoo (for 2 years) and Slackware (for 5 years). The thing I liked about Fedora is frequency of package updates + great community. But lately I have noticed that Fedora is becoming too cutting-edge, nay, bleeding-edge. They changed the DNS client to be strict, without any warning, which broke some of their own packages for two Fedora releases. More critically, their LVM modules are not compatible across Fedora 12, 13, 14 (sometimes). Ubuntu is nicely polished but seems too stable for my liking. Some of the user-space applications are two major version numbers behind (even in testing or unstable or whatever they call it). Is there any Linux distro that has the stability of Ubuntu in kernel space and the bleeding edge in applications (especially harmless applications like, say, Stellarium)?

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  • Fedora 14 - Error 5 - No space on Disk - After installation.

    - by amitahire
    I am new to Fedora, am still figure my way through it. I was going about tweaking it and preparing it to my needs, While installing through yum I dont have much other than the fact that I have to search for the packages. But when I open "Add/remove software" of Fedora. I am greeted with " No space left on the disk" and when I seek for more details it says " Disk Error :[errno 5] Input/output error" I did some research and I saw that it usually occurs during installation, but here I have successfully installed it. Can someone help me with it? Later on I even had notification of the same warning. And mind you I got loads of free space. Thanks for the help

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  • Ubuntu openGL issues

    - by Dank101
    my OpenGL doesn't work at all i get Xlib: extension "GLX" missing on display ":0". lspci output 00:02.0 VGA compatible controller [0300]: Intel Corporation 2nd Generation Core Processor Family Integrated Graphics Controller [8086:0126] (rev 09) (prog-if 00 [VGA controller]) 01:00.0 VGA compatible controller [0300]: NVIDIA Corporation Device [10de:1246] (rev a1) (prog-if 00 [VGA controller]) dmesg | grep -i nvid [ 9.469068] nvidia: module license 'NVIDIA' taints kernel. [ 9.538786] nvidia 0000:01:00.0: power state changed by ACPI to D0 [ 9.538792] nvidia 0000:01:00.0: power state changed by ACPI to D0 [ 9.538796] nvidia 0000:01:00.0: enabling device (0006 -> 0007) [ 9.538803] nvidia 0000:01:00.0: PCI INT A -> GSI 16 (level, low) -> IRQ 16 [ 9.538809] nvidia 0000:01:00.0: setting latency timer to 64 [ 9.538942] NVRM: loading NVIDIA UNIX x86_64 Kernel Module 304.48 Sun Sep 9 [10300.955799] nvidia 0000:01:00.0: restoring config space at offset 0xf (was 0x100, writing 0x10b) [10300.955803] nvidia 0000:01:00.0: restoring config space at offset 0xc (was 0x0, writing 0xfff80000) [10300.955807] nvidia 0000:01:00.0: restoring config space at offset 0x9 (was 0x1, writing 0x4001) [10300.955811] nvidia 0000:01:00.0: restoring config space at offset 0x7 (was 0xc, writing 0xd000000c) [10300.955814] nvidia 0000:01:00.0: restoring config space at offset 0x5 (was 0xc, writing 0xc000000c) [10300.955817] nvidia 0000:01:00.0: restoring config space at offset 0x4 (was 0x0, writing 0xf0000000) [10300.955820] nvidia 0000:01:00.0: restoring config space at offset 0x3 (was 0x800000, writing 0x10) [10300.955823] nvidia 0000:01:00.0: restoring config space at offset 0x1 (was 0x100006, writing 0x100007) my computer is a dell XPS l702x

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  • Shortest-path algorithms which use a space-time tradeoff?

    - by Chris Mounce
    I need to find shortest paths in an unweighted, undirected graph. There are algorithms which can find a shortest path between two nodes, but this can take time. There are also algorithms for computing shortest paths for all pairs of nodes in the graph, but storing such a lookup table would take lots of disk space. What I'm wondering: Is there an algorithm which offers a space-time tradeoff that's somewhere between these two extremes? In other words, is there a way to speed up a shortest-path search, while using less disk space than would be occupied by an all-pairs shortest-path table? I know there are ways to efficiently store lookup tables for this problem, and I already have a couple of ideas for speeding up shortest-path searches using precomputed data. But I don't want to reinvent the wheel if there's already some established algorithm that solves this problem.

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  • How do I write a constant-space length function in Haskell?

    - by Bill
    The canonical implementation of length :: [a] -> Int is: length [] = 0 length (x:xs) = 1 + length xs which is very beautiful but suffers from stack overflow as it uses linear space. The tail-recursive version: length xs = length' xs 0 where length' [] n = n length' (x:xs) n = length xs (n + 1) doesn't suffer from this problem, but I don't understand how this can run in constant space in a lazy language. Isn't the runtime accumulating numerous (n + 1) thunks as it moves through the list? Shouldn't this function Haskell to consume O(n) space and lead to stack overflow? (if it matters, I'm using GHC)

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  • Before the lablel for each of the menu item, there is space with the some color, i want to remove th

    - by anandhinaveen
    Hello, I am using richfaces dropDownMenu component which contains a set of rich menuItems. When the menu items are displayed, a extra space is displayed before the lablel for each of the menu item. But I have a requirement of not displaying the space before the labels and to change the color. I used the css to reduce the space: .rich-menu-item-icon img { width: 0px; } .rich-menu-group-icon img { width: 0px; } but, i need to change the color in that place.

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  • What causes extra space and random alignment in my divs using the 960 grid system?

    - by tomdot
    I have an issue where elements in divs seemingly align randomly, not responding to any align tag, and where divs create extra space above or below its element. I'm using the 960 grid system and I have not altered the stock CSS file lest my fingers blow it up. I've put up a test page here, and please excuse some of the dodgy code :-D It's still in it's very early stages (as am I), but I don't feel I can move on until I can comfortably understand what is causing the issue. The best example on that page is the bottom horizontal bar and elements underneath. The bar adds a few pixels of dead space, and the elements below that align to different parts of their respective divs. Why is it that elements seemingly do not standardise their alignment given no instruction, and what causes extra space again where no instruction was given? My own thought was to relatively position everything individually, but I'm worried this will cause issues and 'break' the grid. Other than that, I unno. Thanks

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  • C++: Why does space always terminate a string when read?

    - by Nullw0rm
    Using type std::string to accept a sentence, for practise (I haven't worked with strings in C++ much) I'm checking if a character is a vowel or not. I got this: for(i = 0; i <= analyse.length(); i++) { if(analyse[i] == 'a' || analyse[i] == 'e' [..etc..]) { ...vowels++; } else { ... ...consenents++; } This works fine if the string is all one word, but the second I add a space (IE: aeio aatest) it will only count the first block and count the space as a consenent, and quit reading the sentence (exiting the for loop or something). Does a space count as no character == null? Or some oddity with std::string?, It would be helpful to know why that is happening!

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  • Efficiency: what block size of kernel-mode memory allocations?

    - by Robert
    I need a big, driver-internal memory buffer with several tens of megabytes (non-paged, since accessed at dispatcher level). Since I think that allocating chunks of non-continuous memory will more likely succeed than allocating one single continuous memory block (especially when memory becomes fragmented) I want to implement that memory buffer as a linked list of memory blocks. What size should the blocks have to efficiently load the memory pages? (read: not to waste any page space) A multiple of 4096? (equally to the page size of the OS) A multiple of 4000? (not to waste another page for OS-internal memory allocation information) Another size? Target platform is Windows NT = 5.1 (XP and above) Target architectures are x86 and amd64 (not Itanium)

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  • How do you know how much space an NSString (using default font) will occupy on a UITableView section

    - by paul_sns
    Basically I want to copy the behavior in iPhone's Calendar where the day of week (Mon, Tue, Wed) is on the left side of the table's section header (left justified) while the formatted date (Apr 1, 2010, May 1, 2010) based on locale is on the right side of the table's section header (right justified). I was thinking of inserting a variable amount of space in between the day of week and formatted date but I first need to know the actual space consumed by the text on both sides to figure out how much space to add. Hope to hear your thoughts on this. Thanks.

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  • Code Golf: Code 39 Bar Code

    - by gwell
    The challenge The shortest code by character count to draw an ASCII representation of a Code 39 bar code. Wikipedia article about Code 39: http://en.wikipedia.org/wiki/Code_39 Input The input will be a string of legal characters for Code 39 bar codes. This means 43 characters are valid: 0-9 A-Z (space) and -.$/+%. The * character will not appear in the input as it is used as the start and stop characters. Output Each character encoded in Code 39 bar codes have nine elements, five bars and four spaces. Bars will be represented with # characters, and spaces will be represented with the space character. Three of the nine elements will be wide. The narrow elements will be one character wide, and the wide elements will be three characters wide. A inter-character space of a single space should be added between each character pattern. The pattern should be repeated so that the height of the bar code is eight characters high. The start/stop character * (bWbwBwBwb) would be represented like this: # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # # # ### ### # ^ ^ ^^ ^ ^ ^ ^^^ | | || | | | ||| narrow bar -+ | || | | | ||| wide space ---+ || | | | ||| narrow bar -----+| | | | ||| narrow space ------+ | | | ||| wide bar --------+ | | ||| narrow space ----------+ | ||| wide bar ------------+ ||| narrow space --------------+|| narrow bar ---------------+| inter-character space ----------------+ The start and stop character * will need to be output at the start and end of the bar code. No quiet space will need to be included before or after the bar code. No check digit will need to be calculated. Full ASCII Code39 encoding is not required, just the standard 43 characters. No text needs to be printed below the ASCII bar code representation to identify the output contents. The character # can be replaced with another character of higher density if wanted. Using the full block character U+2588, would allow the bar code to actually scan when printed. Test cases Input: ABC Output: # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # # # ### ### # ### # # # ### # ### # # ### ### ### # # # # # ### ### # Input: 1/3 Output: # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # # # ### ### # ### # # # ### # # # # # ### ### # # # # # ### ### # Input: - $ (minus space dollar) Output: # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # # # ### ### # # # # ### ### # ### # ### # # # # # # # # ### ### # Code count includes input/output (full program).

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  • how to export bind and keyframe bone poses from blender to use in OpenGL

    - by SaldaVonSchwartz
    EDIT: I decided to reformulate the question in much simpler terms to see if someone can give me a hand with this. Basically, I'm exporting meshes, skeletons and actions from blender into an engine of sorts that I'm working on. But I'm getting the animations wrong. I can tell the basic motion paths are being followed but there's always an axis of translation or rotation which is wrong. I think the problem is most likely not in my engine code (OpenGL-based) but rather in either my misunderstanding of some part of the theory behind skeletal animation / skinning or the way I am exporting the appropriate joint matrices from blender in my exporter script. I'll explain the theory, the engine animation system and my blender export script, hoping someone might catch the error in either or all of these. The theory: (I'm using column-major ordering since that's what I use in the engine cause it's OpenGL-based) Assume I have a mesh made up of a single vertex v, along with a transformation matrix M which takes the vertex v from the mesh's local space to world space. That is, if I was to render the mesh without a skeleton, the final position would be gl_Position = ProjectionMatrix * M * v. Now assume I have a skeleton with a single joint j in bind / rest pose. j is actually another matrix. A transform from j's local space to its parent space which I'll denote Bj. if j was part of a joint hierarchy in the skeleton, Bj would take from j space to j-1 space (that is to its parent space). However, in this example j is the only joint, so Bj takes from j space to world space, like M does for v. Now further assume I have a a set of frames, each with a second transform Cj, which works the same as Bj only that for a different, arbitrary spatial configuration of join j. Cj still takes vertices from j space to world space but j is rotated and/or translated and/or scaled. Given the above, in order to skin vertex v at keyframe n. I need to: take v from world space to joint j space modify j (while v stays fixed in j space and is thus taken along in the transformation) take v back from the modified j space to world space So the mathematical implementation of the above would be: v' = Cj * Bj^-1 * v. Actually, I have one doubt here.. I said the mesh to which v belongs has a transform M which takes from model space to world space. And I've also read in a couple textbooks that it needs to be transformed from model space to joint space. But I also said in 1 that v needs to be transformed from world to joint space. So basically I'm not sure if I need to do v' = Cj * Bj^-1 * v or v' = Cj * Bj^-1 * M * v. Right now my implementation multiples v' by M and not v. But I've tried changing this and it just screws things up in a different way cause there's something else wrong. Finally, If we wanted to skin a vertex to a joint j1 which in turn is a child of a joint j0, Bj1 would be Bj0 * Bj1 and Cj1 would be Cj0 * Cj1. But Since skinning is defined as v' = Cj * Bj^-1 * v , Bj1^-1 would be the reverse concatenation of the inverses making up the original product. That is, v' = Cj0 * Cj1 * Bj1^-1 * Bj0^-1 * v Now on to the implementation (Blender side): Assume the following mesh made up of 1 cube, whose vertices are bound to a single joint in a single-joint skeleton: Assume also there's a 60-frame, 3-keyframe animation at 60 fps. The animation essentially is: keyframe 0: the joint is in bind / rest pose (the way you see it in the image). keyframe 30: the joint translates up (+z in blender) some amount and at the same time rotates pi/4 rad clockwise. keyframe 59: the joint goes back to the same configuration it was in keyframe 0. My first source of confusion on the blender side is its coordinate system (as opposed to OpenGL's default) and the different matrices accessible through the python api. Right now, this is what my export script does about translating blender's coordinate system to OpenGL's standard system: # World transform: Blender -> OpenGL worldTransform = Matrix().Identity(4) worldTransform *= Matrix.Scale(-1, 4, (0,0,1)) worldTransform *= Matrix.Rotation(radians(90), 4, "X") # Mesh (local) transform matrix file.write('Mesh Transform:\n') localTransform = mesh.matrix_local.copy() localTransform = worldTransform * localTransform for col in localTransform.col: file.write('{:9f} {:9f} {:9f} {:9f}\n'.format(col[0], col[1], col[2], col[3])) file.write('\n') So if you will, my "world" matrix is basically the act of changing blenders coordinate system to the default GL one with +y up, +x right and -z into the viewing volume. Then I also premultiply (in the sense that it's done by the time we reach the engine, not in the sense of post or pre in terms of matrix multiplication order) the mesh matrix M so that I don't need to multiply it again once per draw call in the engine. About the possible matrices to extract from Blender joints (bones in Blender parlance), I'm doing the following: For joint bind poses: def DFSJointTraversal(file, skeleton, jointList): for joint in jointList: bindPoseJoint = skeleton.data.bones[joint.name] bindPoseTransform = bindPoseJoint.matrix_local.inverted() file.write('Joint ' + joint.name + ' Transform {\n') translationV = bindPoseTransform.to_translation() rotationQ = bindPoseTransform.to_3x3().to_quaternion() scaleV = bindPoseTransform.to_scale() file.write('T {:9f} {:9f} {:9f}\n'.format(translationV[0], translationV[1], translationV[2])) file.write('Q {:9f} {:9f} {:9f} {:9f}\n'.format(rotationQ[1], rotationQ[2], rotationQ[3], rotationQ[0])) file.write('S {:9f} {:9f} {:9f}\n'.format(scaleV[0], scaleV[1], scaleV[2])) DFSJointTraversal(file, skeleton, joint.children) file.write('}\n') Note that I'm actually grabbing the inverse of what I think is the bind pose transform Bj. This is so I don't need to invert it in the engine. Also note I went for matrix_local, assuming this is Bj. The other option is plain "matrix", which as far as I can tell is the same only that not homogeneous. For joint current / keyframe poses: for kfIndex in keyframes: bpy.context.scene.frame_set(kfIndex) file.write('keyframe: {:d}\n'.format(int(kfIndex))) for i in range(0, len(skeleton.data.bones)): file.write('joint: {:d}\n'.format(i)) currentPoseJoint = skeleton.pose.bones[i] currentPoseTransform = currentPoseJoint.matrix translationV = currentPoseTransform.to_translation() rotationQ = currentPoseTransform.to_3x3().to_quaternion() scaleV = currentPoseTransform.to_scale() file.write('T {:9f} {:9f} {:9f}\n'.format(translationV[0], translationV[1], translationV[2])) file.write('Q {:9f} {:9f} {:9f} {:9f}\n'.format(rotationQ[1], rotationQ[2], rotationQ[3], rotationQ[0])) file.write('S {:9f} {:9f} {:9f}\n'.format(scaleV[0], scaleV[1], scaleV[2])) file.write('\n') Note that here I go for skeleton.pose.bones instead of data.bones and that I have a choice of 3 matrices: matrix, matrix_basis and matrix_channel. From the descriptions in the python API docs I'm not super clear which one I should choose, though I think it's the plain matrix. Also note I do not invert the matrix in this case. The implementation (Engine / OpenGL side): My animation subsystem does the following on each update (I'm omitting parts of the update loop where it's figured out which objects need update and time is hardcoded here for simplicity): static double time = 0; time = fmod((time + elapsedTime),1.); uint16_t LERPKeyframeNumber = 60 * time; uint16_t lkeyframeNumber = 0; uint16_t lkeyframeIndex = 0; uint16_t rkeyframeNumber = 0; uint16_t rkeyframeIndex = 0; for (int i = 0; i < aClip.keyframesCount; i++) { uint16_t keyframeNumber = aClip.keyframes[i].number; if (keyframeNumber <= LERPKeyframeNumber) { lkeyframeIndex = i; lkeyframeNumber = keyframeNumber; } else { rkeyframeIndex = i; rkeyframeNumber = keyframeNumber; break; } } double lTime = lkeyframeNumber / 60.; double rTime = rkeyframeNumber / 60.; double blendFactor = (time - lTime) / (rTime - lTime); GLKMatrix4 bindPosePalette[aSkeleton.jointsCount]; GLKMatrix4 currentPosePalette[aSkeleton.jointsCount]; for (int i = 0; i < aSkeleton.jointsCount; i++) { F3DETQSType& lPose = aClip.keyframes[lkeyframeIndex].skeletonPose.joints[i]; F3DETQSType& rPose = aClip.keyframes[rkeyframeIndex].skeletonPose.joints[i]; GLKVector3 LERPTranslation = GLKVector3Lerp(lPose.t, rPose.t, blendFactor); GLKQuaternion SLERPRotation = GLKQuaternionSlerp(lPose.q, rPose.q, blendFactor); GLKVector3 LERPScaling = GLKVector3Lerp(lPose.s, rPose.s, blendFactor); GLKMatrix4 currentTransform = GLKMatrix4MakeWithQuaternion(SLERPRotation); currentTransform = GLKMatrix4TranslateWithVector3(currentTransform, LERPTranslation); currentTransform = GLKMatrix4ScaleWithVector3(currentTransform, LERPScaling); GLKMatrix4 inverseBindTransform = GLKMatrix4MakeWithQuaternion(aSkeleton.joints[i].inverseBindTransform.q); inverseBindTransform = GLKMatrix4TranslateWithVector3(inverseBindTransform, aSkeleton.joints[i].inverseBindTransform.t); inverseBindTransform = GLKMatrix4ScaleWithVector3(inverseBindTransform, aSkeleton.joints[i].inverseBindTransform.s); if (aSkeleton.joints[i].parentIndex == -1) { bindPosePalette[i] = inverseBindTransform; currentPosePalette[i] = currentTransform; } else { bindPosePalette[i] = GLKMatrix4Multiply(inverseBindTransform, bindPosePalette[aSkeleton.joints[i].parentIndex]); currentPosePalette[i] = GLKMatrix4Multiply(currentPosePalette[aSkeleton.joints[i].parentIndex], currentTransform); } aSkeleton.skinningPalette[i] = GLKMatrix4Multiply(currentPosePalette[i], bindPosePalette[i]); } Finally, this is my vertex shader: #version 100 uniform mat4 modelMatrix; uniform mat3 normalMatrix; uniform mat4 projectionMatrix; uniform mat4 skinningPalette[6]; uniform lowp float skinningEnabled; attribute vec4 position; attribute vec3 normal; attribute vec2 tCoordinates; attribute vec4 jointsWeights; attribute vec4 jointsIndices; varying highp vec2 tCoordinatesVarying; varying highp float lIntensity; void main() { tCoordinatesVarying = tCoordinates; vec4 skinnedVertexPosition = vec4(0.); for (int i = 0; i < 4; i++) { skinnedVertexPosition += jointsWeights[i] * skinningPalette[int(jointsIndices[i])] * position; } vec4 skinnedNormal = vec4(0.); for (int i = 0; i < 4; i++) { skinnedNormal += jointsWeights[i] * skinningPalette[int(jointsIndices[i])] * vec4(normal, 0.); } vec4 finalPosition = mix(position, skinnedVertexPosition, skinningEnabled); vec4 finalNormal = mix(vec4(normal, 0.), skinnedNormal, skinningEnabled); vec3 eyeNormal = normalize(normalMatrix * finalNormal.xyz); vec3 lightPosition = vec3(0., 0., 2.); lIntensity = max(0.0, dot(eyeNormal, normalize(lightPosition))); gl_Position = projectionMatrix * modelMatrix * finalPosition; } The result is that the animation displays wrong in terms of orientation. That is, instead of bobbing up and down it bobs in and out (along what I think is the Z axis according to my transform in the export clip). And the rotation angle is counterclockwise instead of clockwise. If I try with a more than one joint, then it's almost as if the second joint rotates in it's own different coordinate space and does not follow 100% its parent's transform. Which I assume it should from my animation subsystem which I assume in turn follows the theory I explained for the case of more than one joint. Any thoughts?

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  • Can compressing Program Files save space *and* give a significant boost to SSD performance?

    - by Christopher Galpin
    Considering solid-state disk space is still an expensive resource, compressing large folders has appeal. Thanks to VirtualStore, could Program Files be a case where it might even improve performance? Discovery In particular I have been reading: SSD and NTFS Compression Speed Increase? Does NTFS compression slow SSD/flash performance? Will somebody benchmark whole disk compression (HD,SSD) please? (may have to scroll up) The first link is particularly dreamy, but maybe head a little too far in the clouds. The third link has this sexy semi-log graph (logarithmic scale!). Quote (with notes): Using highly compressable data (IOmeter), you get at most a 30x performance increase [for reads], and at least a 49x performance DECREASE [for writes]. Assuming I interpreted and clarified that sentence correctly, this single user's benchmark has me incredibly interested. Although write performance tanks wretchedly, read performance still soars. It gave me an idea. Idea: VirtualStore It so happens that thanks to sanity saving security features introduced in Windows Vista, write access to certain folders such as Program Files is virtualized for non-administrator processes. Which means, in normal (non-elevated) usage, a program or game's attempt to write data to its install location in Program Files (which is perhaps a poor location) is redirected to %UserProfile%\AppData\Local\VirtualStore, somewhere entirely different. Thus, to my understanding, writes to Program Files should primarily only occur when installing an application. This makes compressing it not only a huge source of space gain, but also a potential candidate for performance gain. Testing The beginning of this post has me a bit timid, it suggests benchmarking NTFS compression on a whole drive is difficult because turning it off "doesn't decompress the objects". However it seems to me the compact command is perfectly capable of doing so for both drives and individual folders. Could it be only marking them for decompression the next time the OS reads from them? I need to find the answer before I begin my own testing.

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  • How to detect which Space the user is on in Mac OS X Leopard?

    - by georgebrock
    Mac OS X Leopard has a virtual desktop implementation called Spaces. I want to programatically detect which space the user is currently on. Cocoa is preferable but AppleScript is acceptable if there's no other way. I've seen a couple of AppleScript implementations, but the techniques they used seemed a bit too hacky to use in production code (one relied on causing and error and then parsing the error message to get the current space, the other interrogated the Spaces menu GUI)

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  • Java - How to force resize JCheckBox to prevent clicking the empty space ?

    - by Brad
    When i create a JCheckBox in my Swing application i leave some extra space after its label, so if the JCheckBox label is for example 100 pixels width, i make the JCheckBox 120 pixels for safety. The problem as at runtime, it's not nice that a user can click on the empty space after the JCheckBox label and it can be actually clicked, like this : I wonder if there is a way to resize the JCheckBox at runtime to exactly fit the text inside it, depending on the font type/size used ? This seems fancy a bit, but i like to make things look perfect :)

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  • How do you avoid an invalid search space in a genetic algorithm?

    - by Dave
    I am developing a GA for a school project and I've noticed that upon evaluating my functions for fitness, an individual is equivalent to its inverse. For example, the set (1, 1, -1, 1) is equivalent to (-1, -1, 1, -1). To shrink my search space and reach a solution more efficiently, how can I avoid my crossovers from searching in this second half of the search space?

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