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1 G (x, y) 2 2 1 G (x, y) 2 2

scientific: 10e+01 fixed decimal: 10

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (1004 / 1065) [2001-3-29 11:32:15]

S , (x, y) , S , (x, y) .

However, EAP was specifically designed to be a protocol with a simple message set to carry exchanges of short messages for authentication purposes The fact that the EAP message set only includes 4 messages (EAP request, response, success and failure messages) attests to this For those reasons, EAP is not suitable to transport large amounts of data traffic and hence lacks sophisticated transport capabilities, such as packet ordering Therefore, when run over link layer, the link layer protocol must provide EAP with packet reordering services Another transport issue with EAP is that EAP is a lock-step protocol, when it comes to request/response pairs In other words, when a request is sent from the authenticator to the peer, another request cannot be sent until a response is received from that peer Note that this is different from retransmissions needed in case of transport loss, for which EAP has provided provisions.

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If we should wish to have the "e" of the scientific notation output display as "E", we use the uppercase manipulator To revert back to lowercase, we use the nouppercase manipulator (The uppercase manipulator does not cause all alpha characters to be displayed in uppercase!) By default, the overloaded input operators skip white space (blank, tab, newline, formfeed, and carriage return) Given the sequence

char ch; while ( cin >> ch ) //

(5.4) (5.5)

executes four times to read the characters a through d, skipping the intervening blanks, possible tabs, and newline characters The manipulator noskipws causes the input operator not to skip white space:

char ch; cin >> noskipws; while ( cin >> ch ) // cin >> skipws;

Figure 5.3a shows the 3D representation of a Gabor lter (either real or imaginary) at = 0.554 radians, = 0 radians, and = 10, and Figures 5.3b and 5.3c show an image with the real and imaginary parts of the complex lter. In order to extract a Gabor feature at a location (x, y) of an image I, the real and imaginary parts of the lter are applied separately to the same location in the image, and a magnitude is computed from the two results. Thus, the Gabor lter coef cient at a location (x, y) in an image I with a Gabor lter , is given by C (x, y) = (I(x, y) {

The while loop now requires seven iterations to read the characters a through d To return to the default behavior, the skipws manipulator is applied to cin When we write

cout "please enter a value: ";

the literal string is stored in a buffer associated with cout There are a number of conditions that cause the buffer to be flushed that is, emptied in our example by writing it to standard output: 1 The buffer can become full In this case, it must be flushed to read the next value 2 We can flush the buffer explicitly using either the flush, ends, or endl manipulator:

2 , (x, y)})

On the other hand, the EAP success and failure messages are not acknowledged and hence provisions must be made for lost success and failure messages, if so required The final transport issue that needs consideration is fragmentation Fragmentation typically needs to be performed when the maximum transmission unit (MTU) size is smaller than the protocol packet size When an EAP authentication mechanism uses large packet sizes (such as packets carrying digital certificates), fragmentation may be required Take an example where the MTU size is around 1400 bytes, while a certificate chain consisting of 3 certificates each around 1 KByte is to be sent This means the original packets would need to be divided into multiple fragments.

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (1005 / 1065) [2001-3-29 11:32:15]

char ch[2]; ch[0] = 'a'; ch[1] = 'b'; cout ch ends; // inserts a newline then flushes the buffer cout "hi!" endl;

+ (I(x, y) {

3 unitbuf, an internal stream state variable, can be set to empty the buffer after each output operation 4 The ostream object can be tied to an istream, in which case the ostream buffer is flushed whenever the istream reads from the input stream cout is predefined to be tied to cin:

cintie( &cout );

2 , (x, y)}) .

cin >> ival;

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