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SystemNet is a large namespace that contains many members It is far beyond the scope of this chapter to discuss them all or to discuss all aspects related to Internet programming (In fact, an entire book is needed to fully cover networking in detail) However, it is worthwhile to list the members of SystemNet so you have an idea of what is available for your use

Part II:

The classes defined by SystemNet are shown here:

( 1)

AuthenticationManager Cookie CookieContainer CredentialCache DnsEndPoint DnsPermissionAttribute DownloadProgressChangedEventArgs EndPoint FileWebRequest FtpWebRequest HttpListener HttpListenerContext HttpListenerPrefixCollection HttpListenerResponse HttpWebRequest IPAddress IPEndPointCollection IrDAEndPoint OpenReadCompletedEventArgs ProtocolViolationException ServicePointManager SocketPermission TransportContext UploadFileCompletedEventArgs UploadStringCompletedEventArgs WebClient WebHeaderCollection WebPermissionAttribute WebRequest WebRequestMethodsFile WebRequestMethodsHttp WebUtility Authorization CookieCollection CookieException Dns DnsPermission DownloadDataCompletedEventArgs DownloadStringCompletedEventArgs EndpointPermission FileWebResponse FtpWebResponse HttpListenerBasicIdentity HttpListenerException HttpListenerRequest HttpVersion HttpWebResponse IPEndPoint IPHostEntry NetworkCredential OpenWriteCompletedEventArgs ServicePoint SocketAddress SocketPermissionAttribute UploadDataCompletedEventArgs UploadProgressChangedEventArgs UploadValuesCompletedEventArgs WebException WebPermission WebProxy WebRequestMethods WebRequestMethodsFtp WebResponse

26:

a & c is false a | b is true a | c is true Use short-circuit && and || a && b is true a && c is false a || b is true a || c is true

.

Let s look closely at how the & and | are implemented They are shown here:

1 3

// Overload | for short-circuit evaluation public static ThreeD operator |(ThreeD op1, ThreeD op2) { if( ((op1x != 0) || (op1y != 0) || (op1z != 0)) | ((op2x != 0) || (op2y != 0) || (op2z != 0)) ) return new ThreeD(1, 1, 1); else return new ThreeD(0, 0, 0); } // Overload & for short-circuit evaluation public static ThreeD operator &(ThreeD op1, ThreeD op2) { if( ((op1x != 0) && (op1y != 0) && (op1z != 0)) & ((op2x != 0) && (op2y != 0) && (op2z != 0)) ) return new ThreeD(1, 1, 1); else return new ThreeD(0, 0, 0); }

Notice first that both now return an object of type ThreeD Pay attention to how this object is generated If the outcome of the operation is true, then a true ThreeD object (one in which at least one coordinate is non-zero) is created and returned If the outcome is false, then a false object is created and returned Thus, in a statement like this

if(a & b) ConsoleWriteLine("a & b is true"); else ConsoleWriteLine("a & b is false");

A solid has base in the x-y plane consisting of a unit square with center at the origin and vertices on the axes The vertical cross section at position x is itself a square Refer to Figure 89 What is the volume of this solid

the outcome of a & b is a ThreeD object, which in this case is a true object Since the operators true and false are defined, this resulting object is subjected to the true operator, and a bool result is returned In this case, the result is true and the if succeeds Because the necessary rules have been followed, the short-circuit operators are now available for use on ThreeD objects They work like this The first operand is tested by using operator true (for ||) or operator false (for &&) If it can determine the outcome of the operation, then the corresponding & or | is not evaluated Otherwise, the corresponding overloaded & or | is used to determine the result Thus, using a && or || causes the corresponding & or | to be invoked only when the first operand cannot determine the outcome of the expression For example, consider this statement from the program:

9:

if(a || c) ConsoleWriteLine("a || c is true");

The true operator is first applied to a Since a is true in this situation, there is no need to use the | operator method However, if the statement were rewritten like this:

It is sufficient to calculate the volume of the right half of this solid, and to double the answer Of course the extent of x is then 0 x 1/ 2 At position x, the height of the upper edge of the square base is 1/ 2 x So the base of the vertical square slice is 2( 1/ 2 x) (Figure 810) The area of the slice is then A( x) = [2( 1/ 2 x) ]2 = ( 2 2x) 2