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A pointer with a value of 0 is not addressing an object Applying the member access operator to a 0-valued pointer is always troublesome The first logical AND operator guards against that happening Equally troublesome is an out-of-bounds array index The second and third operands guard against that possibility The last operand is safe to evaluate only when the first three operands evaluate as true The logical NOT operator (!) evaluates to true if its operand has a value of false or zero; otherwise, it evaluates to false For example:

bool found = false; // while the item has not been found // and ptr still addresses an object while ( ! found && ptr ) { found = lookup( *ptr ); ++ptr; }

1

RADIUS Access Request/ EAP response Pass-through RADIUS Access Accept: Cipher suite, Key material EAP Success

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (145 / 1065) [2001-3-29 11:32:04]

evaluates to true as long as found is equal to false It is a shorthand notation for the explicit test:

(1.32)

if ( found )

is a shorthand notation for the explicit test:

1 = n

if ( found == true )

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Although use of the binary relational operators, such as the less-than or inequality operator, is straightforward, there is one potential pitfall we must be aware of: The order of evaluation of the left and right operands is deliberately left undefined by both the C and the C++ Standards and must therefore not contain order dependencies For example, in the following

(1.33)

// oops! language does not define order of evaluation if ( ia[ index++ ] < ia[ index ] ) // swap elements

Figure 10.9 Client CHAP authentication through EAP-TTLS messaging Note: Brackets indicate data that is encrypted using TLS record layer and is invisible to the NAS.

the programmer presumes that the left operand is evaluated first and that therefore ia[0] is compared to be less than ia[1] The language, however, does not guarantee a left-to-right evaluation order; an implementation may in fact evaluate the right-hand operand first, in which case ia[0] is compared to itself and ia[1] is never evaluated One safe and portable reimplementation looks like this:

if ( ia[ index ] < ia[ index+1 ] ) // swap elements ++index;

where cj is the centroid of the jth class and c is the global centroid in the feature space. Similar to the linear case, the transformation G of KDA can be computed by solving the following optimization problem: G = arg max trace

A second potential pitfall is the following Our intention is to determine whether ival, jval, and kval are each unique values Do you see what's wrong

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (146 / 1065) [2001-3-29 11:32:04]

(1.34)

// oops! this does not determine if the 3 values are unequal if ( ival != jval != kval ) // do something

As we've implemented it, the associated values 0, 1, and 0 cause the expression to evaluate as true The reason is that the left operand of the second inequality expression is the true/false result of the first that is, kval is tested for inequality against the promoted integer values of 0 or 1 To accomplish our test, we might rewrite the expression as follows:

The TLS handshake phase is performed as described earlier The end of this phase is marked in Figure 109 To spice things up, here we have shown an example of how EAP-TTLS can be used for cipher suite negotiation and key distribution for the secure communications between the client and the NAS (see the last assumption in the section describing EAP-TTLS functional elements) As shown, both client and NAS can indicate their preferences to the TTLS server and let that server act as the mediator This explains what is shown in the picture as the cipher suite suggestion added by the NAS to the RADIUS access request following the EAP-Identity request/response pair The client, on the other hand, waits for the establishment of the TLS record layer before sending its cipher suite preferences to the TLS server (as explained below).

It follows from the Representer Theorem [47] that columns of G lie in the span of the images of training data in the feature space. That is, G = (X)B, for some matrix B Rn (k 1) , where (X) = [ (x1 ), . . . , (xn )] is the data matrix in the feature space. Substituting Eq. (1.35) into Eq. (1.34), we can obtain the matrix B by solving the following optimization problem: B = arg max trace

Exercise 44 Which, if any, of the following are likely to be incorrect or not portable Why How might each be corrected (Note that the type of the object(s) is not significant in these examples)

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