Search Results

Search found 10662 results on 427 pages for 'cursor position'.

Page 59/427 | < Previous Page | 55 56 57 58 59 60 61 62 63 64 65 66  | Next Page >

  • How do you implement position-sensitive zooming inside a JScrollPane?

    - by tucuxi
    I am trying to implement position-sensitive zooming inside a JScrollPane. The JScrollPane contains a component with a customized 'paint' that will draw itself inside whatever space it is allocated - so zooming is as easy as using a MouseWheelListener that resizes the inner component as required. But I also want zooming into (or out of) a point to keep that point as central as possible within the resulting zoomed-in (or -out) view (this is what I refer to as 'position-sensitive' zooming), similar to how zooming works in google maps. I am sure this has been done many times before - does anybody know the "right" way to do it under Java Swing?. Would it be better to play with Graphic2D's transformations instead of using JScrollPanes? Sample code follows: package test; import java.awt.*; import java.awt.event.*; import java.awt.geom.*; import javax.swing.*; public class FPanel extends javax.swing.JPanel { private Dimension preferredSize = new Dimension(400, 400); private Rectangle2D[] rects = new Rectangle2D[50]; public static void main(String[] args) { JFrame jf = new JFrame("test"); jf.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); jf.setSize(400, 400); jf.add(new JScrollPane(new FPanel())); jf.setVisible(true); } public FPanel() { // generate rectangles with pseudo-random coords for (int i=0; i<rects.length; i++) { rects[i] = new Rectangle2D.Double( Math.random()*.8, Math.random()*.8, Math.random()*.2, Math.random()*.2); } // mouse listener to detect scrollwheel events addMouseWheelListener(new MouseWheelListener() { public void mouseWheelMoved(MouseWheelEvent e) { updatePreferredSize(e.getWheelRotation(), e.getPoint()); } }); } private void updatePreferredSize(int n, Point p) { double d = (double) n * 1.08; d = (n > 0) ? 1 / d : -d; int w = (int) (getWidth() * d); int h = (int) (getHeight() * d); preferredSize.setSize(w, h); getParent().doLayout(); // Question: how do I keep 'p' centered in the resulting view? } public Dimension getPreferredSize() { return preferredSize; } private Rectangle2D r = new Rectangle2D.Float(); public void paint(Graphics g) { super.paint(g); g.setColor(Color.red); int w = getWidth(); int h = getHeight(); for (Rectangle2D rect : rects) { r.setRect(rect.getX() * w, rect.getY() * h, rect.getWidth() * w, rect.getHeight() * h); ((Graphics2D)g).draw(r); } } }

    Read the article

  • Finding N contiguous zero bits in an integer to the left of the MSB position of another integer

    - by James Morris
    The problem is: given an integer val1 find the position of the highest bit set (Most Significant Bit) then, given a second integer val2 find a contiguous region of unset bits, with the minimum number of zero bits given by width to the left of the position (ie, in the higher bits). Here is the C code for my solution: typedef unsigned int t; unsigned const t_bits = sizeof(t) * CHAR_BIT; _Bool test_fit_within_left_of_msb( unsigned width, t val1, t val2, unsigned* offset_result) { unsigned offbit = 0; unsigned msb = 0; t mask; t b; while(val1 >>= 1) ++msb; while(offbit + width < t_bits - msb) { mask = (((t)1 << width) - 1) << (t_bits - width - offbit); b = val2 & mask; if (!b) { *offset_result = offbit; return true; } if (offbit++) /* this conditional bothers me! */ b <<= offbit - 1; while(b <<= 1) offbit++; } return false; } Aside from faster ways of finding the MSB of the first integer, the commented test for a zero offbit seems a bit extraneous, but necessary to skip the highest bit of type t if it is set. I have also implemented similar algorithms but working to the right of the MSB of the first number, so they don't require this seemingly extra condition. How can I get rid of this extra condition, or even, are there far more optimal solutions? Edit: Some background not strictly required. The offset result is a count of bits from the high bit, not from the low bit as maybe expected. This will be part of a wider algorithm which scans a 2D array for a 2D area of zero bits. Here, for testing, the algorithm has been simplified. val1 represents the first integer which does not have all bits set found in a row of the 2D array. From this the 2D version would scan down which is what val2 represents. Here's some output showing success and failure: t_bits:32 t_high: 10000000000000000000000000000000 ( 2147483648 ) --------- ----------------------------------- *** fit within left of msb test *** ----------------------------------- val1: 00000000000000000000000010000000 ( 128 ) val2: 01000001000100000000100100001001 ( 1091569929 ) msb: 7 offbit:0 + width: 8 = 8 mask: 11111111000000000000000000000000 ( 4278190080 ) b: 01000001000000000000000000000000 ( 1090519040 ) offbit:8 + width: 8 = 16 mask: 00000000111111110000000000000000 ( 16711680 ) b: 00000000000100000000000000000000 ( 1048576 ) offbit:12 + width: 8 = 20 mask: 00000000000011111111000000000000 ( 1044480 ) b: 00000000000000000000000000000000 ( 0 ) offbit:12 iters:10 ***** found room for width:8 at offset: 12 ***** ----------------------------------- *** fit within left of msb test *** ----------------------------------- val1: 00000000000000000000000001000000 ( 64 ) val2: 00010000000000001000010001000001 ( 268469313 ) msb: 6 offbit:0 + width: 13 = 13 mask: 11111111111110000000000000000000 ( 4294443008 ) b: 00010000000000000000000000000000 ( 268435456 ) offbit:4 + width: 13 = 17 mask: 00001111111111111000000000000000 ( 268402688 ) b: 00000000000000001000000000000000 ( 32768 ) ***** mask: 00001111111111111000000000000000 ( 268402688 ) offbit:17 iters:15 ***** no room found for width:13 ***** (iters is the count of iterations of the inner while loop)

    Read the article

  • How to set webview scroll position within it on initial launch ?

    - by mob-king
    I have an issue with Webview to set the initial scroll position within it when first launched. By default it is left , I want it to scroll and show the view from center. Or is it possible to resize the view to fit the screen width initially, instead of scroll bars. I need this since the left part of the page I am loading is blank making user feel the page is not loaded.

    Read the article

  • How to preserve sibling element position when one sibling is absolutely positioned?

    - by Casey
    In the snippet below, the child div is normally positioned until it is :hovered , when it becomes absolutely positioned. The reasoning behind this markup is to simulate a popup style in a limited environment where I can't use a <select> (among other limitations). When child is hovered, the sibling elements jump around, which is expected, as the contents of the block have changed. But how can I preserve their positioning? That is, what CSS can I add to prevent the siblings from jumping around when child is hovered. Javascript is also not allowed, so please no answers using JS. HTML: <div class="container"> <div class="child"> <span class="d4"></span> <label><input type="radio" name="radio" value="1"/>One</label> <label><input type="radio" name="radio" value="2"/>Two</label> </div> <input type="text" name="sibling"/> <button name="sibling2">Button</button> </div> CSS: .container, .child, button { display:inline-block; } .child { vertical-align: middle; width: 35px; height: 35px; } .child:hover { background: gray; position:absolute; width: 100px; height: auto; } .child:hover > .d4 { display: none; } .child label { display:none; } .child:hover label { display: inline-block; } .d4 { background-position: -411px -1px; width: 35px; height: 35px; background-image: url("https://i.imgur.com/zkgyBOi.png"); background-repeat: no-repeat; color: transparent; display: inline-block; } Here's a fiddle: http://jsfiddle.net/cpctZ/1/

    Read the article

  • What variable dictates position of non-focused elements in the roundabout plugin?

    - by kristina childs
    Part of the problem here is that i'm not sure what the best language to use in order to find the solution. I search and searched so please forgive if this is already a thread somewhere. I'm using the roundabout plugin to cycle through 3 divs. Each div is 794px wide, which makes the roundabout-in-focus element 794 and the two not in focus 315.218px wide, positioned so half of each is hidden by the in-focus div. This is all well and good, however the total width of the display needs to stay within 1000px (ideally 980px, but i can fudge if need be.) Basically I want to make the non-focused divs be 3/4 hidden by the in-focus div but for the life of me can't figure out what variables i need to edit in order to do it. Unfortunately it's not one of the many easily-changed options like z-index and minScale. i tried minScale but it's clear this isn't going to work. the plugin outputs this code: <li class="roundabout-moveable-item" style="position: absolute; left: -57px; top: 205px; width: 319.982px; height: 149.513px; opacity: 0.7; z-index: 146; font-size: 5.6px;"> i need to find out what changes the left positioning so it's shifted closer to the center of the stage, like this: <li class="roundabout-moveable-item" style="position: absolute; left: 5px; top: 205px; width: 319.982px; height: 149.513px; opacity: 0.7; z-index: 146; font-size: 5.6px;"> i tried playing with the positioning functions of the plugin but all that did was shift everything in tandem left or right. any help is greatly appreciated. this site is going to be awesome once i figure out all this jquery stuff! here is a link to my .js file: http://avalon.eaw.com/scripts/jquery.roundabout2.js i've got an overflow:hidden on the to help guide the positioning of those no-focused items.

    Read the article

  • elisp: posn-at-point returns nil after goto-char. How to update the display before posn-at-point?

    - by Cheeso
    In emacs lisp, posn-at-point is documented as: posn-at-point is a built-in function in C source code. (posn-at-point &optional POS WINDOW) . Return position information for buffer POS in WINDOW. POS defaults to point in WINDOW; WINDOW defaults to the selected window. . Return nil if position is not visible in window. Otherwise, the return value is similar to that returned by event-start for a mouse click at the upper left corner of the glyph corresponding to the given buffer position: (WINDOW AREA-OR-POS (X . Y) TIMESTAMP OBJECT POS (COL . ROW) IMAGE (DX . DY) (WIDTH . HEIGHT)) The posn- functions access elements of such lists. ok, now I've got a function that looks something like this: (defun my-move-and-popup-menu () "move the point, then pop up a menu." (goto-char xxxx) (setq p (posn-at-point)) (my-popup-menu p ...) ) Basically, move the point, then retrieve the screen position at that point, and then popup a menu at that screen position. But I am finding that posn-at-point returns non-nil, only if the xxxx character position (the after position) is visible in the window, before the call to goto-char. It seems that the position is not actually updated until exit from the function. If goto-char goes a long way, more than one screenful, then the retrieved position is always nil, and my code doesn't know where to popup the menu. The reason I suggest that the position is not actually updated until exit from the function - when the menu successfully pops up, the cursor is clearly visible in its previous location while the popup menu is being displayed. When I dismiss the menu, the cursor moves to where I expected it to move, after the goto-char call. How can I get the position to be really updated, between goto-char and posn-at-point, so that posn-at-point will not return nil? In a Windows Forms application I would call Form.Update() or something similar to update the display in the middle of an event handler. What's the emacs version of that?

    Read the article

  • Sorting a list of numbers with modified cost

    - by David
    First, this was one of the four problems we had to solve in a project last year and I couldn’t find a suitable algorithm so we handle in a brute force solution. Problem: The numbers are in a list that is not sorted and supports only one type of operation. The operation is defined as follows: Given a position i and a position j the operation moves the number at position i to position j without altering the relative order of the other numbers. If i j, the positions of the numbers between positions j and i - 1 increment by 1, otherwise if i < j the positions of the numbers between positions i+1 and j decreases by 1. This operation requires i steps to find a number to move and j steps to locate the position to which you want to move it. Then the number of steps required to move a number of position i to position j is i+j. We need to design an algorithm that given a list of numbers, determine the optimal (in terms of cost) sequence of moves to rearrange the sequence. Attempts: Part of our investigation was around NP-Completeness, we make it a decision problem and try to find a suitable transformation to any of the problems listed in Garey and Johnson’s book: Computers and Intractability with no results. There is also no direct reference (from our point of view) to this kind of variation in Donald E. Knuth’s book: The art of Computer Programing Vol. 3 Sorting and Searching. We also analyzed algorithms to sort linked lists but none of them gives a good idea to find de optimal sequence of movements. Note that the idea is not to find an algorithm that orders the sequence, but one to tell me the optimal sequence of movements in terms of cost that organizes the sequence, you can make a copy and sort it to analyze the final position of the elements if you want, in fact we may assume that the list contains the numbers from 1 to n, so we know where we want to put each number, we are just concerned with minimizing the total cost of the steps. We tested several greedy approaches but all of them failed, divide and conquer sorting algorithms can’t be used because they swap with no cost portions of the list and our dynamic programing approaches had to consider many cases. The brute force recursive algorithm takes all the possible combinations of movements from i to j and then again all the possible moments of the rest of the element’s, at the end it returns the sequence with less total cost that sorted the list, as you can imagine the cost of this algorithm is brutal and makes it impracticable for more than 8 elements. Our observations: n movements is not necessarily cheaper than n+1 movements (unlike swaps in arrays that are O(1)). There are basically two ways of moving one element from position i to j: one is to move it directly and the other is to move other elements around i in a way that it reaches the position j. At most you make n-1 movements (the untouched element reaches its position alone). If it is the optimal sequence of movements then you didn’t move the same element twice.

    Read the article

< Previous Page | 55 56 57 58 59 60 61 62 63 64 65 66  | Next Page >