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At point A substituting h h = 0091 h 4 2p h w h (314) = 225 h yA = = 80 b p 180 2p h w 2 yA = A2 = ( b )2 yA = y1 = = 2p h (314) = 3170 h 2 80 p 180

Use pointer arithmetic *(p+0): 0 *(p+1): 1 *(p+2): 2 *(p+3): 3 *(p+4): 4 *(p+5): 5 *(p+6): 6 *(p+7): 7 *(p+8): 8 *(p+9): 9

As the program illustrates, a pointer expression with this general form *(ptr + i) can be rewritten using array-indexing syntax like this: ptr[i] There are two important things to understand about indexing a pointer: First, no boundary checking is applied Thus, it is possible to access an element beyond the end of the array to which the pointer refers Second, a pointer does not have a Length property So, using the pointer, there is no way of knowing how long the array is

By using some algebraic manipulations, we can reduce a variety of indeterminate limits to expressions which can be treated by l H pital s Rule We explore some of these techniques in this section

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The displacement of part 2, the parabolic curve, in Sec 26 y2 = vA Substituting, we obtain y2 = h w q 2 1 2p h w 2 q 2 + b w 2 ( b )2 w 2 h q 2 h pq 2 = + 2 b (b ) 1 h = + h p = 1285 h 4 2 q2 w 40 p 180 = 933 h 314

Although strings are implemented as objects in C#, it is possible to access the characters in a string through a pointer To do so, you will assign a pointer to the start of the string to a char* pointer using a fixed statement like this: xed(char* p = str) { // After the fixed statement executes, p will point to the start of the array of characters that make up the string This array is null-terminated, which means that it ends with a zero You can use this fact to test for the end of the array Null-terminated character arrays are the way that strings are implemented in C/C++ Thus, obtaining a char* pointer to a string allows you to operate on strings in much the same way as does C/C++ Here is a program that demonstrates accessing a string through a char* pointer:

// Use fixed to get a pointer to the start of a string using System; class FixedString { unsafe static void Main() { string str = "this is a test";

Substituting, 0091 h + 1285 h + 1199 h = 2 h = 07767 in Let us now nd the displacements: y1 = 0091( 07767) = 0071 in y2 = 1285( 07767) = 0998 in y3 = 1191( 07767) = 0896 in

Part I:

x

// Point p to start of str fixed(char* p = str) { // Display the contents of str via p for(int i=0; p[i] != 0; i++) ConsoleWrite(p[i]); } ConsoleWriteLine(); } }

Substituting to nd the velocity vA = 225( 07767) = 175 ips vB = 933( 07767) = 725 ips vT = maximum velocity = vB + DvB to T = 725 + 175 = 900 ips Also, the maximum acceleration AA = 3170( 07767) = 2460 in sec 2 The curves may now be plotted in Fig 36b If they were to be compared with a trapezoidal acceleration curve, we would nd that this curve has a slightly lower maximum acceleration and the advantage of lower required cutting accuracy in the initial and nal rise portions Also, the vibrations induced at high speeds should be slightly smaller than those of the trapezoidal curve

The output is shown here:

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