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Let s go through a quick refresher on state machines so we all start from the same place Essentially, a state machine is based on the concept of conditions ( states ) and transitions A condition is a set of circumstances that indicate the current status or situation of the process being modeled Events occur and cause a transition from one condition to another Unlike sequential workflows, there is no defined path through the workflow Instead, the path taken by the workflow is determined by the events that occur as the workflow is processing A state machine can only be in one condition at a time State machines are good at modeling both complex processes as well as simpler process They excel at modeling nondeterministic processes that is, processes that do not have a prescribed path that is known at design time.

Finally, due to their nature and the inherent event-driven paradigm, state machines are naturally suited to modeling processes that progress based on the occurrence of a trappable event from a collection of events known and planned for at design time..

as extra input for a post-processing effect. This way, although you re processing a 2D image, you know the distance between each pixel and the camera. You can use this, for example, to blur only the parts of the image that are farther away from the camera, while leaving the sections close to the camera sharp!

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In the following code, you can find the method that triggers a fly-by action All it needs are the beginning and final camera positions and targets: private void InitBezier(Vector3 startPosition, Vector3 startTarget, Vector3 endPosition, Vector3 endTarget) { bezStartPosition = startPosition; bezEndPosition = endPosition; bezMidPosition = (bezStartPosition + bezEndPosition) / 20f; Vector3 midShiftDirecton = new Vector3(1, 1, 0)*2; Vector3 cameraDirection = endPosition - startPosition; Vector3 perpDirection = Vector3Cross(upV, cameraDirection); perpDirectionNormalize(); Vector3 midShiftDirecton = new Vector3(0, 1, 0) + perpDirection; bezMidPosition += cameraDirectionLength() * midShiftDirecton; bezStartTarget = startTarget; bezEndTarget = endTarget; bezTime = 00f; } All the startPosition, endPosition, startTarget, and endTarget arguments can immediately be stored in their corresponding global variables The midPosition, which is an extra position you will insert between the starting and ending positions to bend the curve, needs to be calculated.

When linking to custom subtotals at the bottom of a group, the automatically generated GetPivotData formulas return a #REF! error. This is a bug that has existed in Excel for several versions, but the formula can be easily corrected. To fix the formula so it calculates correctly, remove the Data, from the formula. The corrected formula for this example is =GETPIVOTDATA($A$3,"Category[Bars;Average]")

Console.WriteLine("--- Left-shift Operator ---"); Console.WriteLine("Before: {0} ({1})", ConvertToBinary(b), b); Console.WriteLine("After: {0} ({1})", ConvertToBinary(result), result); b = 60; // right shift three places result = b >> 3; // show the before and after values Console.WriteLine("\n--- Right-shift Operator ---"); Console.WriteLine("Before: {0} ({1})", ConvertToBinary(b), b); Console.WriteLine("After: {0} ({1})", ConvertToBinary(result), result); // wait for input before exiting Console.WriteLine("Press enter to finish"); Console.ReadLine(); } private static string ConvertToBinary(int value) { return String.Format("{0:00000000}", int.Parse(Convert.ToString(value, 2))); } } The code in Listing 5-21 shifts a value two places to the left and shifts another value three places to the right. Compiling and running the code in Listing 5-21 produces the following results: --- Left-shift Operator --Before: 00001111 (15) After: 00111100 (60) --- Right-shift Operator --Before: 00111100 (60) After: 00000111 (7) Press enter to finish You can see that when I shift to the left, all of the 1s in the binary string are moved two places to the left. This has the effect of changing the value that is represented from 15 to 60 (see Figure 5-1 for details of why). Notice that two new zeros have been put in to the rightmost positions. When I shift three places to the right, the bits are moved in the opposite direction notice that one of the 1s is shifted out of the byte entirely. When this happens, the 1 value is gone for good; shifting back to the left doesn t restore it. Notice also that when a value is shifted out, it doesn t reappear on the other side (as though the values were in a loop); the positions are always filled with 0 values.

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