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places into p the memory address of the variable num This address is the computer s internal location of the variable It has nothing to do with the value of num The operation of the & can be remembered as returning the address of Therefore, the preceding assignment statement could be read as p receives the address of num For example, assume the variable num uses memory location 2000 to store its value Also assume that num has a value of 100 Then, after the preceding assignment, p will have the value 2000 The second operator, *, is the complement of & It is a unary operator that returns the value of the variable located at the address that follows For example, if p contains the memory address of the variable num,

q = *p;

The growth of Mid-Band Ethernet is driven by multiple converging needs: the universal adoption of the Internet Protocol (IP), the economics of Ethernet, and the cost and complexity of real-world fiber deployments

StrokeThickness="2" RadiusX="10" RadiusY="10"/> <ContentPresenter HorizontalAlignment="Center" VerticalAlignment="Center"/> <Rectangle x:Name="buttonTopShine" Grid.ColumnSpan="3" RadiusX="10" RadiusY="10"> <Rectangle.Fill> <LinearGradientBrush StartPoint= "0,0" EndPoint="0,1" Opacity="0.8"> <GradientStop Color="White" Offset="0"/> <GradientStop Color="Transparent" Offset="0.3"/> </LinearGradientBrush> </Rectangle.Fill> </Rectangle> </Grid>

places the value of num into q Following through with this example, q has the value 100 because 100 is stored at location 2000, which is the memory address that was stored in p The operation of the * can be remembered as at address In this case the statement could be read as q receives the value at address p

The following program illustrates the foregoing discussion:

The XAML mirrors the diagram from figure 83 The glow is accomplished by creating a rounded rectangle the same size as the button b but with a bright yellow color (#FFEE08) This hex color code is somewhat brighter than just Yellow, albeit harder to read The next element is a dark object c Again, it s another rounded rectangle, but instead of being a solid color, it uses a radial gradient that starts out black and then goes to transparent By having two gradient stops that are black, we get a slightly bigger dark area before the gradient drops off We also set the origin of the object at 09,09, putting it toward the lower-right corner You can play with all these numbers to try and get slightly different effects Next is our button d.

Enterprises continue to adopt more and more IP-based applications and dramatically grow their consumption of packet network capacity Business applications such as file sharing, training, storage networks, and video conferencing are all growing in coverage and bandwidth requirements Voice over IP (VoIP) is just starting to replace analog voice as the primary mechanism for telephony All carriers are in the midst of rolling out more and more VoIP applications and removing their dependence on traditional voice services On the residential side, VoIP is also a driving application, as is IP television (IPTV) With IPTV, consumers can watch digital high-definition video over their broadband connection This triple-play of services is the goal of every carrier All of these applications have similar requirements high bandwidth, high reliability, and highly controlled QoS with low latency, loss, and jitter Ethernet provides the technology that has enabled all of these applications in the LAN, and with Mid-Band Ethernet, across the WAN as well

#include <stdioh> int main(void) { int num, q; int *p; num = 100; /* num is assigned 100 */ p = # /* p receives num's address */ q = *p; /* q is assigned num's value indirectly through p */ printf("%d", q); /* prints 100 */ return 0; }

The preceding program displays the value 100 Unfortunately, the multiplication sign and the at address sign are the same, and the bitwise AND and the address of sign are the same These operators have no relationship to each other Both & and * have a higher precedence than the binary arithmetic operators You must make sure that your pointer variables always point to the correct type of data For example, when you declare a pointer to be of type int, the compiler assumes that any address it holds points to an integer value Because C allows you to assign any address to a pointer variable, the following code fragment compiles (although C++ Builder will issue a warning message) but does not produce the desired result