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Search found 1210 results on 49 pages for 'positive'.

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  • 2D Side scroller collision detection

    - by Shanon Simmonds
    I am trying to do some collision detection between objects and tiles, but the tiles do not have there own x and y position, they are just rendered to the x and y position given, there is an array of integers which has the ids of the tiles to use(which are given from an image and all the different colors are assigned different tiles) int x0 =

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  • How do i approach this collision model?

    - by PeeS
    this is the game level prototype i have already implemented. It has few objects per room to allow me to finally add some collision detection/response code into it. VIDEO As you can probably see, every object inside has it's own AABB, even the room itself has AABB. So a player is like 'inside the Room AABB'. My player will be exactly

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  • WCF REST Error Handler

    - by Elton Stoneman
    I’ve put up on GitHub a sample WCF error handler for REST services, which returns proper HTTP status codes in response to service errors.   The code is very simple – a ServiceBehavior implementation which can be specified in config to tag the RestErrorHandler to a service. Any uncaught exceptions will be routed to the error handler,

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  • Smoothing rotation

    - by Lewis
    I've spent the last three days trying to work out how to rotate a sprite smoothly depending on the velocity.x value of the sprite. I'm using this: float Proportion = 9.5; float maxDiff = 200; float rotation = fmaxf(fminf(playerVelocity.x * Proportion, maxDiff), -maxDiff); player.rotation = rotation; The behaviour is what I required but

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  • Poll on Entity Framework 4 &ndash; one year on

    - by Eric Nelson
    12 months back (today is March 15th 2010) on the 16th of  March 2009 I created a poll on Entity Framework v1 – the marmite of ORMs? A quick poll…. Entity Framework v1 was getting a mixed reception at the time – I met developers who genuinely hated it and I met developers who were loving the productivity improvements they were seeing.

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  • When you’re on a high, start something big

    - by BuckWoody
    Most days are pretty average – we have some highs, some lows, and just regular old work to do. But some days the sun is shining, your co-workers are especially nice, and everything just falls into place. You really *enjoy* what you do. Don’t let that moment pass. All of us have “big” projects that we need to tackle. Things that are

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  • Choosing a VS project type (C++)

    - by typoknig
    Hi all, I do not use C++ much (I try to stick to the easier stuff like Java and VB.NET), but the lately I have not had a choice. When I am picking a project type in VS for some C++ source I download, what project type should I pick? I had just been sticking with Win32 Console Applications, but I just downloaded some code (below)

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  • Calculate year for end date: PostgreSQL

    - by Dave Jarvis
    Background Users can pick dates as shown in the following screen shot: Any starting month/day and ending month/day combinations are valid, such as: Mar 22 to Jun 22 Dec 1 to Feb 28 The second combination is difficult (I call it the "tricky date scenario") because the year for the ending month/day is before the year for

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  • Having two ODP.NET (ODAC) versions in the same server

    - by vizcaynot
    Hello: Some months ago, a colleague of mine installed ODAC 11.106.21 in a server using XCOPY and then he developed many applications that use this client without problems (in test and production windows servers). Past week, I developed an application under ODAC 11.1.07.20. When I asked him to install these new ODAC version

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  • I have a having a a matrix index bounds in matlab

    - by Ben Fossen
    I keep getting the error( this is in Matlab) Attempted to access r(0,0); index must be a positive integer or logical. Error in == Romberg at 15 I ran it with Romberg(1.3, 2.19,8) I think the problem is the statment is not logical because I made it positive and still got the same error. anyone got some ideas of what i

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  • Help Understanding Function

    - by Fred F.
    What does the following function perform? public static double CleanAngle(double angle) { while (angle < 0) angle += 2 * System.Math.PI; while (angle > 2 * System.Math.PI) angle -= 2 * System.Math.PI; return angle; } This is how it is used with ATan2. I

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  • RSA Factorization problem

    - by dada
    At class we found this programming problem, and currently, we have no idea how to solve it. The positive integer n is given. It is known that n = p * q, where p and q are primes, p<=q and |q-k*p|<10^5 for some given positive integer k. You must find p and q. Input: 35 1 121 1 1000730021 9 Output: 5

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  • How to move a kinect skeleton to another position

    - by Ewerton
    I am working on a extension method to move one skeleton to a desired position in the kinect field os view. My code receives a skeleton to be moved and the destiny position, i calculate the distance between the received skeleton hip center and the destiny position to find how much to move, then a iterate in the

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  • Unsigneds in order to prevent negative numbers

    - by Bruno Brant
    let's rope I can make this non-sujective Here's the thing: Sometimes, on fixed-typed languages, I restrict input on methods and functions to positive numbers by using the unsigned types, like unsigned int or unsigned double, etc. Most libraries, however, doesn't seem to think that way. Take C#

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  • eliminating noise/spikes

    - by tgv
    I have a measurement data with similar positive and negative values which should be like: ReqData=[0 0 -2 -2 -2 -2 -2 -2 0 0 0 -2 -2 -2 -2 0 0 2 2 2 2 2 2 0 0 2 2 2 2 2 0 0 2 2 2 2 2 0 0 2 2 2 0 0]' However, there are some measurement noises in the data - so the real data is like this:

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  • Prove 2^sqrt(log(n))= O(n^a)

    - by user1830621
    I posted a question similar to this earlier, although it seemed like it was much easier. I know the underlying principle of Big-O notation says that to prove that 2^sqrt(log(n)) is O(n^a), there must exist a positive value c for which c(n^a) = 2^sqrt(log(n)) for all values n = N. c*n^a

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