Am I invalidating the entire screen on every call?
android, invalidation, ondraw, optimization
Solution
Kraken
Q: But what about the logs, surely if my loops are getting executed it means that all the canvas.draw will be getting executed too? A: Yes, the whole drawing will be executed in your sample code. You have to optimize the rendering process by yourself, in `onDraw` method.
Q: How does the system know, what piece of code will "only" redraw the dirty area? A: `Canvas::getClipBounds` will give you a dirty rect, which you should draw something on. Inside your `for loop` in `onDraw`, compare the dirty rect with the rect which you want to draw. Then do `continue` if they do not intersect.
But remember, if you have several area set to dirty, the returned rect will be a union of all dirty areas. Please see the following two questions below: Getting the dirty region inside draw() Android: invalidate(dirty)
Hope this will help you.
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The author is right. But this still can be optimized.
Calling `invalidate(Rect)` will automatically set a clip area for the canvas. (That's why `canvas.getClipBounds()` can return that area). Then, during `onDraw()`, anything drawing out of the clip area, will be ignored. They do not appear on the screen, so it REALLY reduce the time of drawing. But ignoring them still costs overhead. Therefore, for graphical intensive app, `onDraw()` could be better optimized if you exclude them in advance.
You can find a great example for optimizing `onDraw()` in android's `KeyboardView`, which provide the view of your android's input method. http://grepcode.com/file/repository.grepcode.com/java/ext/com.google.android/android/1.5_r4/android/inputmethodservice/KeyboardView.java
Problem
I am new to Android Development and reading the book `Hello Android`. It uses a Sudoku example, and the code that I am referring to is here. In this , onTouchScreen, it calls select method, that calls invalidate twice. The question is that, on `invalidating` is the `onDraw` method called right after that? So will in this case, inside my select method, it will do - invalidate - call onDraw - Do some stuff - invalidate - call onDraw Is this how it will happen, also, will the entire screen be regenerated? All the numbers and hints etc., because from the book the author says In an earlier version of this example, I invalidated the entire screen whenever the cursor was moved. Thus, on every key press, the whole puzzle had to be redrawn. This caused it to lag noticeably. Switching the code to invalidate only the smallest rectangles that changed made it run much faster. What exactly is he trying to say here? Added Info I added some logs in the onDraw method, some at the starting, some in the for loop. Whenever I touched a new rectangle, all the logs were called. Doesnt that mean that the entire screen is geting repopulated, since all the code in onDraw is reexecuted?