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  • Detecting coincident subset of two coincident line segments

    - by Jared Updike
    This question is related to: How do I determine the intersection point of two lines in GDI+? (great explanation of algebra but no code) How do you detect where two line segments intersect? (accepted answer doesn't actually work) But note that an interesting sub-problem is completely glossed over in most solutions which just return null for the coincident case even though there are three sub-cases: coincident but do not overlap touching just points and coincident overlap/coincident line sub-segment For example we could design a C# function like this: public static PointF[] Intersection(PointF a1, PointF a2, PointF b1, PointF b2) where (a1,a2) is one line segment and (b1,b2) is another. This function would need to cover all the weird cases that most implementations or explanations gloss over. In order to account for the weirdness of coincident lines, the function could return an array of PointF's: zero result points (or null) if the lines are parallel or do not intersect (infinite lines intersect but line segments are disjoint, or lines are parallel) one result point (containing the intersection location) if they do intersect or if they are coincident at one point two result points (for the overlapping part of the line segments) if the two lines are coincident

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  • Inner shadow issue in Illustrator CS5

    - by Joe Conlin
    I gave a comp to my client that I did in Photoshop. I used an inner shadow but now have realized the in Illustrator CS5 I have no such "easy" filter. I have spent 2 days seaching the web, trying tutorials, etc. to no avail. Every tutorial seems to use text but I am not using text. Anyone that can answer I would forever been in debt... :) This is the image with the inner shadow inside the stripes that I am needing to duplicate. Thanks!

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  • Oracle 10g multiple DELETE statements

    - by bmw0128
    I'm building a dml file that first deletes records that may be in the table, then inserts records. Example: DELETE from foo where field1='bar'; DELETE from foo where fields1='bazz'; INSERT ALL INTO foo(field1, field2) values ('bar', 'x') INTO foo(field1, field2) values ('bazz', 'y') SELECT * from DUAL; When I run the insert statement by itself, it runs fine. When I run the deletes, only the last delete runs. Also, it seems to be necessary to end the multiple insert with the select, is that so? If so, why is that necessary? In the past, when using MySQL, I could just list multiple delete and insert statements, all individually ending with a semicolon, and it would run fine.

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  • How to create lines with Athens?

    - by Kilon
    I have no clue how to create lines with Athens. I took a look at Cairo docs but I cant see how Athens is related to Cairo. http://zetcode.com/gfx/cairo/basicdrawing/ In the above link I cant find any equivalent for cairo_set_line_width(cr, 1); I tried to look inside Athens but is nowhere to be found. Overall I find the Athens architecture quite confusing though Cairo looks simple. Any idea how to makes this work ?

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  • Cannot save Adobe Illustrations as .pdf files

    - by Parastar
    Hi, I use adobe Illustrator creative suite 3. Now there seems to be a problem. I can't save my Illustrations In the .pdf format, since this format seems to preserve the graphic quality this Is vitally Important to me. If I save the Graphic In .jpef /.gif or png there Is some serious loss of quality so I need to find out a way to get this sorted. when I try to save the graphic as pdf It says "An unknown error has occured" please can anyone help me out here? Thank you,

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  • JavaScript - Multiple signals received after click

    - by Angelo A
    I'm creating a webapp using jQueryMobile. When I'm using the app and I click a button it runs the script multiple times. For example: I have a submit button: <input type="submit" id="login-normal" value="Login" /> And I have this JavaScript for debugging on which this error occurs: $("input#login-normal").live('click',function() { console.log("Test"); }); On the very first click it works (and it goes to another screen for example), but when I go back to that screen and I click again, it outputs multiple console.logs

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  • PDF Making with Rotated Image for iPhone

    - by MobiHunterz
    I'm having problems with drawing rotated images on PDF, my output is worse. My case is, we don't know have any fixed co-ordinates. X,Y, rotation, etc. depends on ImageView itself. I select the ImageView and rotate it through Sliders. Check on ZOSH application. I need to implement functionalities like that app. I want to make PDF by adding images one by one. Please send me link for any example that can help me out, I'm stuck here. I'm drawing the image on PDF based on center of the imageview. Please help me, Thank You.

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  • Associating multiple MySQL queries w/ PHP?

    - by anon
    I am trying to create a simple inventory request system. A user can enter multiple SKU's that will query the inventory database. My problem is I am trying to do is associate these multiple queries into a type of list. This list can later be retrieved and contains all queries that were submitted simultaneously. When the list is filled it can then be deleted. Is this possible? I am just looking to be pointed in the right direction. Thanks

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  • Cloning ID3DXMesh with declration that has 12 floats breaks?

    - by meds
    I have the following vertex declration: struct MESHVERTInstanced { float x, y, z; // Position float nx, ny, nz; // Normal float tu, tv; // Texcoord float idx; // index of the vertex! float tanx, tany, tanz; // The tangent const static D3DVERTEXELEMENT9 Decl[6]; static IDirect3DVertexDeclaration9* meshvertinstdecl; }; And I declare it as such: const D3DVERTEXELEMENT9 MESHVERTInstanced::Decl[] = { { 0, 0, D3DDECLTYPE_FLOAT3, D3DDECLMETHOD_DEFAULT, D3DDECLUSAGE_POSITION, 0 }, { 0, 12, D3DDECLTYPE_FLOAT3, D3DDECLMETHOD_DEFAULT, D3DDECLUSAGE_NORMAL, 0 }, { 0, 24, D3DDECLTYPE_FLOAT2, D3DDECLMETHOD_DEFAULT, D3DDECLUSAGE_TEXCOORD, 0 }, { 0, 32, D3DDECLTYPE_FLOAT1, D3DDECLMETHOD_DEFAULT, D3DDECLUSAGE_TEXCOORD, 1 }, { 0, 36, D3DDECLTYPE_FLOAT3, D3DDECLMETHOD_DEFAULT, D3DDECLUSAGE_TANGENT, 0 }, D3DDECL_END() }; What I try to do next is copy an ID3DXMesh into another one with the new vertex declaration as such: model->CloneMesh( model->GetOptions(), MESHVERTInstanced::Decl, gd3dDevice, &pTempMesh ); When I try to get the FVF size of pTempMesh (D3DXGetFVFVertexSize(pTempMesh-GetFVF())) I get '0' though the size should be 48. The whole thing is fine if I don't have the last declaration, '{ 0, 36, D3DDECLTYPE_FLOAT3, D3DDECLMETHOD_DEFAULT, D3DDECLUSAGE_TANGENT, 0 },' in it and the CloneMesh function does not return a FAIL. I've also tried using different declarations such as D3DDECLUSAGE_TEXCOORD and that has worked fine, returning a size of 48. Is there something specific about D3DDECLUSAGE_TANGENT I don't know? I'm at a complete loss as to why this isn't working...

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  • Transform shape built of contour splines to simple polygons

    - by Cheery
    I've dumped glyphs from truetype file so I can play with them. They have shape contours that consist from quadratic bezier curves and lines. I want to output triangles for such shapes so I can visualize them for the user. Traditionally I might use libfreetype or scan-rasterise this kind of contours. But I want to produce extruded 3D meshes from the fonts and make other distortions with them. So, how to polygonise shapes consisting from quadratic bezier curves and lines? There's many contours that form the shape together. Some contours are additive and others are subtractive. The contours are never open. They form a loop. (Actually, I get only contour vertices from ttf glyphs, those vertices define whether they are part of the curve or not. Even though it is easy to decompose these into bezier curves and lines, knowing the data is represented this way may be helpful for polygonizing the contours to triangles)

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  • Jquery Accordion and multiple slideshows

    - by Dipesh Parmar
    I've been using a lot of slideshows recently on my sites and one thing thats puzzled me is using more than one slideshow per page. I'm currently working on my own site, experimenting with Jquery Accordion. I've managed to adopt a very simple javascript slideshow, see below: http://dvpwebdesign.com/test/accordion/blank.html However i'm unable to either incorporate or use a different multiple slideshow plugin. I dont need slideshow navigation, so the The Cycle plugin works really well and i know you can use multiple slideshows. But if i either use Cycle alongside the current javascript slideshow, or only use the Cycle slideshow to avoid any possible conflict, the Accordion menu stops working. I just cant see what i am doing wrong, can anyone help?

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  • Merging and splitting overlapping rectangles to produce non-overlapping ones

    - by uj
    I am looking for an algorithm as follows: Given a set of possibly overlapping rectangles (All of which are "not rotated", can be uniformly represented as (left,top,right,bottom) tuplets, etc...), it returns a minimal set of (non-rotated) non-overlapping rectangles, that occupy the same area. It seems simple enough at first glance, but prooves to be tricky (at least to be done efficiently). Are there some known methods for this/ideas/pointers? Methods for not necessarily minimal, but heuristicly small, sets, are interesting as well, so are methods that produce any valid output set at all.

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  • Can a JPEG compressed image be rotated without a loss in quality?

    - by Mat
    JPEG is a lossy compression scheme, so decompression-manipulation-recompression normally reduces the image quality further for each step. Is it possible to rotate a JPEG image without incurring further loss? From what little I know of the JPEG algorithm, it naively seems possible to avoid further loss with a bit of effort. Which common image manipulation programs (e.g. GIMP, Paint Shop Pro, Windows Photo Gallery) and graphic libraries cause quality loss when performing a rotation and which don't?

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  • How to get ImageButton size within Android GridView?

    - by wufoo
    I'm subclassing ImageButton in order to draw lines on it and trying to figure out where the actual button coordinates are within my gridview. I am using onGlobalLayout to setup Top, Bottom, Right and Left, but these seem to be for the actual "square" within the grid, and not the actual button (see image). The purple lines are drawn in myImageButton.onDraw() using coords gathered from myImageButton.onGlobalLayout(). I thought these would be for the button, but they seem to be from something else. Not sure what. I'd like the purple lines to match the outline of the button so the lines I draw appear on the button and not just floating out in the LinearLayout somewhere. The light blue is the background color of the vertical LinearLayout holding the Textview (for the number) and myImageButton. Any way to get the actual button size? XML Layout: <FrameLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/lay_cellframe" android:layout_width="fill_parent" android:layout_height="fill_parent" > <LinearLayout android:layout_width="fill_parent" android:layout_height="fill_parent" android:gravity="fill_vertical|fill_horizontal" android:orientation="vertical" > <TextView android:id="@+id/tv_cell" android:layout_width="fill_parent" android:layout_height="wrap_content" android:layout_margin="2dp" android:gravity="center" android:text="TextView" android:textSize="10sp" /> <com.example.icaltest2.myImageButton android:id="@+id/imageButton1" android:layout_width="fill_parent" android:layout_height="fill_parent" android:layout_gravity="center" android:layout_margin="0dp" android:adjustViewBounds="false" android:background="@android:drawable/btn_default" android:scaleType="fitXY" android:src="@android:color/transparent" /> </LinearLayout> </FrameLayout> myImageButton.java public myImageButton (Context context, AttributeSet attrs) { super (context, attrs); mBounds = new Rect(); ViewTreeObserver vto = this.getViewTreeObserver (); vto.addOnGlobalLayoutListener (ogl); Log.d (TAG, "myImageButton"); } ... OnGlobalLayoutListener ogl = new OnGlobalLayoutListener() { @Override public void onGlobalLayout () { Rect b = getDrawable ().getBounds (); mBtnTop = b.centerY () - (b.height () / 2); mBtnBot = b.centerY () + (b.height () / 2); mBtnLeft = b.centerX () - (b.width () / 2); mBtnRight = b.centerX () + (b.width () / 2); } }; ... @Override protected void onDraw (Canvas canvas) { super.onDraw (canvas); Paint p = new Paint (); p.setStyle (Paint.Style.STROKE); p.setStrokeWidth (1); p.setColor (Color.MAGENTA); canvas.drawCircle (mBtnLeft, mBtnTop, 2, p); canvas.drawCircle (mBtnLeft, mBtnBot, 2, p); canvas.drawCircle (mBtnRight, mBtnTop, 2, p); canvas.drawCircle (mBtnRight, mBtnBot, 2, p); canvas.drawRect (mBtnLeft, mBtnTop, mBtnRight, mBtnBot, p); }

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  • how to dynamically recolor a CGGradientRef

    - by saintmac
    I have three CGGradientRef's that I need to be able to dynamically recolor. When I Initialise the CGGradientRef's the first time I get the expected result, but every time I attempt to change the colors nothing happens. Why? gradient is an instance variable ins a subclass of CALayer: @interface GradientLayer : CALayer { CGGradientRef gradient; //other stuff } @end Code: if (gradient != NULL) { CGGradientRelease(gradient); gradient = NULL; } RGBA color[360]; //set up array CGColorSpaceRef rgb = CGColorSpaceCreateDeviceRGB(); gradient = CGGradientCreateWithColorComponents ( rgb, color, NULL, sizeof (color) / sizeof (color[0]) ); CGColorSpaceRelease(rgb); [self setNeedsDisplay];

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  • identify the input that is multiple of 11 and odd or even java

    - by Bolor Ch
    i am trying to write code to determine the nature of input using if statement only. The nature of input could be following: a multiple of 11 even or odd. For the code below, when I enter my input, it does not display the result as "input:NOT:ODD". Also how can I check multiple conditions with if statement? (else is not considered) import java.util.Scanner; public class test { public static void main( String args[] ) { Scanner input = new Scanner( System.in ); int x; int EO; int Mult; System.out.print ( "Enter value: " ); x = input.nextInt(); EO = x % 2; Mult = x % 11; if (EO > 0 && Mult > 0) { System.out.printf ("%d:NOT:ODD"); } } }

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  • How to Create a Multiple Question ?

    - by user537422
    Hi Guys, I need to retrieve 6 questions from a plist and check the answer if is correct from the plist itself?? i will use a QR code scanner api to scan for answer, the api will covert to a string and read from the plist to check if the answer is correct... is there any tutorial or references for me to look @ ?? In my plist there is: question ~ Dictionary with the following strings: NumberOfOption ~ which define if the question is a multiple choice or a QR code question Question ~ question itself Answer ~ Answer itself Option 1 ~ 4 ~ if it is a multiple choice question Thanks in advance for answering my questions, i appreciated it cheers Desmond

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  • Why are my lines getting thicker and thicker?

    - by mystify
    I try to draw some lines with different colors. This code tries to draw two rectangles with 1px thin lines. However, the second rectangle is drawn with 2px width lines, while the first one is drawn with 1px width. - (void)addLineFrom:(CGPoint)p1 to:(CGPoint)p2 context:(CGContextRef)context { // set the current point CGContextMoveToPoint(context, p1.x, p1.y); // add a line from the current point to the wanted point CGContextAddLineToPoint(context, p2.x, p2.y); } - (void)drawRect:(CGRect)rect { CGContextRef context = UIGraphicsGetCurrentContext(); CGPoint from, to; // ----- draw outer black frame (left, bottom, right) ----- CGContextBeginPath(context); // set the color CGFloat lineColor[4] = {26.0f * colorFactor, 26.0f * colorFactor, 26.0f * colorFactor, 1.0f}; CGContextSetStrokeColor(context, lineColor); // left from = CGPointZero; to = CGPointMake(0.0f, rect.size.height); [self addLineFrom:from to:to context:context]; // bottom from = to; to = CGPointMake(rect.size.width, rect.size.height); [self addLineFrom:from to:to context:context]; // right from = to; to = CGPointMake(rect.size.width, 0.0f); [self addLineFrom:from to:to context:context]; CGContextStrokePath(context); CGContextClosePath(context); // ----- draw the middle light gray frame (left, bottom, right) ----- CGContextSetLineWidth(context, 1.0f); CGContextBeginPath(context); // set the color CGFloat lineColor2[4] = {94.0f * colorFactor, 94.0f * colorFactor, 95.0f * colorFactor, 1.0f}; CGContextSetStrokeColor(context, lineColor2); // left from = CGPointMake(200.0f, 1.0f); to = CGPointMake(200.0f, rect.size.height - 2.0f); [self addLineFrom:from to:to context:context]; // bottom from = to; to = CGPointMake(rect.size.width - 2.0f, rect.size.height - 2.0f); [self addLineFrom:from to:to context:context]; // right from = to; to = CGPointMake(rect.size.width - 2.0f, 1.0f); [self addLineFrom:from to:to context:context]; // top from = to; to = CGPointMake(1.0f, 1.0f); [self addLineFrom:from to:to context:context]; CGContextStrokePath(context); }

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  • opengl: question about glutMainLoop()

    - by lego69
    can somebody explain how does glutMainLoop work? and second question, why glClearColor(0.0f, 0.0f, 1.0f, 1.0f); defined after glutDisplayFunc(RenderScene); cause firstly we call glClear(GL_COLOR_BUFFER_BIT); and only then define glClearColor(0.0f, 0.0f, 1.0f, 1.0f); int main(int argc, char* argv[]) { glutInit(&argc, argv); glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB); glutInitWindowSize(800, 00); glutInitWindowPosition(300,50); glutCreateWindow("GLRect"); glutDisplayFunc(RenderScene); glutReshapeFunc(ChangeSize); glClearColor(0.0f, 0.0f, 1.0f, 1.0f); <-- glutMainLoop(); return 0; } void RenderScene(void) { // Clear the window with current clearing color glClear(GL_COLOR_BUFFER_BIT); // Set current drawing color to red // R G B glColor3f(1.0f, 0.0f, 1.0f); // Draw a filled rectangle with current color glRectf(0.0f, 0.0f, 50.0f, -50.0f); // Flush drawing commands glFlush(); }

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  • How can I draw the control points of a Bézier Path in Java?

    - by Sanoj
    I have created a Path of Bézier curves and it works fine to draw the path. But I don't know How I can draw the Control Points together with the Path. Is that possible or do I have to keep track of them in another datastructure? I am creating the path with: Path2D.Double path = new Path2D.Double(); path.moveTo(0,0); path.curveTo(5, 6, 23, 12, 45, 54); path.curveTo(34, 23, 12, 34, 2, 3); And drawing it with: g2.draw(path);

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  • Draw arrow on line

    - by Pete
    Hi, I have this code: CGPoint arrowMiddle = CGPointMake((arrowOne.x + arrowTo.x)/2, (arrowOne.y + arrowTo.y)/2); CGPoint arrowLeft = CGPointMake(arrowMiddle.x-40, arrowMiddle.y); CGPoint arrowRight = CGPointMake(arrowMiddle.x, arrowMiddle.y + 40); [arrowPath addLineToScreenPoint:arrowLeft]; [arrowPath addLineToScreenPoint:arrowMiddle]; [arrowPath addLineToScreenPoint:arrowRight]; [[mapContents overlay] addSublayer:arrowPath]; [arrowPath release]; with this output: http://yfrog.com/edschermafbeelding2010032p What have i to add to get the left and right the at same degree of the line + 30°. If someone has the algorithm of drawing an arrow on a line, pleas give it. It doesn't matter what programming language it is... Thanks

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  • Calculating the angle between two points

    - by kingrichard2005
    I'm currently developing a simple 2D game for Android. I have a stationary object that's situated in the center of the screen and I'm trying to get that object to rotate and point to the area on the screen that the user touches. I have the constant coordinates that represent the center of the screen and I can get the coordinates of the point that the user taps on. I'm using the formula outlined in this forum: How to get angle between two points? -It says as follows "If you want the the angle between the line defined by these two points and the horizontal axis: double angle = atan2(y2 - y1, x2 - x1) * 180 / PI;". -I implemented this, but I think the fact the I'm working in screen coordinates is causing a miscalculation, since the Y-coordinate is reversed. I'm not sure if this is the right way to go about it, any other thoughts or suggestions are appreciated.

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