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A Single Path for the Customer s Data Imagine for a moment an EVC connecting eight sites, spread across a continent, providing the equivalent of a single shared medium to the customer Then assume, further, that this customer s data traffic patterns are highly variable, with high-volume unicast and multicast streams originating and being consumed at different sites at different moments in time, as the customer sees fit Question: Does the provider actually want to carry that customer s traffic along the most direct path from site to site Figure 1311 shows such a network, simplified to only four customer connections C1 C4 Assuming that the most direct paths between the various customer sites are as shown by the dotted arrows in Figure 1311, you can see that this one customer s data is scattered all over the network In fact, there are only three links, T-V, T-W, and V-Y, that never carry data for this customer And even those links would be used if some sort of equal-cost multipathing were used in the network While such diffusion is often desirable in an enterprise network, in order to spread the load as much as possible in a carrier network, such advantages can easily be outweighed by the added difficulty of ensuring quality of service guarantees, determining how much bandwidth is required on any given link, and debugging data-dependent customer problems In Figure 1312, this customer s EVC is assumed to be confined to a single spanning tree rooted at bridge W Only the physical links shown as solid lines are used to carry the EVC Compared to Figure 1311, traffic between one pair of customer sites, C1-C3, takes a longer path through the network However, only this one pair suffers; the other five possible pairings take optimal paths The amount of bandwidth required by the customer on any given link can reasonably be set by the provider, and debugging problems is simpler More importantly, if you assume that the network shown is a provider.

You can also create objects when defining a class by putting the variable names after the closing curly brace, in exactly the same way as you do with a structure

The general form of a class declaration is class class-name { private data and functions public: public data and functions } object-list; Of course, the object-list may be empty Inside the declaration of queue, prototypes to the member functions are used In C++, when you need to tell the compiler about a function, you must use its full prototype form Further, in C++, all functions must be prototyped Prototypes are not optional, as they are in C When it comes time to actually code a function that is a member of a class, you must tell the compiler to which class the function belongs For example, here is one way to code the qput( ) function:

private static string dictionaryToUse = "";

bridge (8021ad) network, only bridge W in Figure 1312 learns this customer s MAC addresses All the other bridges, S-V and X-Z, have only two ports each that carry this customer s data, and hence they do not need to learn any customer MAC addresses By contrast, in Figure 1311, every bridge except bridge V has at least three ports carrying this customer s data, and hence all but that bridge must learn at least some of this customer s MAC addresses Thus, you can see by this comparison that having a single spanning tree for a given customer can be, on balance, an improvement over the perfect routing of data

void queue::qput(int i) { if(sloc==99) { cout << "Queue is full\n"; return; } sloc++; q[sloc] = i; }

The :: is called the scope resolution operator Essentially, it tells the compiler that this version of qput( ) belongs to the queue class or, put differently, that this qput( ) is in queue s scope In C++, several different classes can use the same function names The compiler knows which function belongs to which class because of the scope resolution operator and the class name To call a member function from a part of your program that is not a member of the class, you must use an object s name and the dot operator For example, this fragment calls init( ) for object a:

protected void OnSelectDictionary(object sender, RoutedEventArgs e) { SettingsPageFunction pageFunction = new SettingsPageFunction(dictionaryToUse); pageFunction.Return += new ReturnEventHandler<String>( OnSettingsPageFunctionReturned);

queue a, b; ainit();