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(2x(PmodQ) + { assume r = P mod Q } ib)modQ . That is, r = P mod Q is maintained invariant by taking m to be (2xr + b) mod Q. This still involves a 'mod Q' operation, which we must eliminate. The strategy is to split the assignment to r into the composition of two assignments, the first assigning 2xr + b to r, and the second assigning r mod Q to r, as shown below. A r - PmodQ } P,r \- 2xP + b,2xr + b ; { (KP A rmodQ = PmodQ } r := r m o d Q A r = PmodQ } . This simplifies the task to replacing the second assignment by something not involving the 'mod Q' operation. The assignment r := r m o d Q may be replaced by skip if the precondition r = r mod Q holds. Now, r = r mod Q is equivalent to O^r < Q . Also, by an easy calculation, (15.37)



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Listing 253: Looking for Command-Line Options (Simplified)

private static void CustomWritePropertiesAsElements( SerializationContext serializationContext, Versioned element, XmlWriter writer) { Versioned instance = element as Versioned; if (!serializationContextResultFailed && instance != null) { String[] parts = { instanceMajorToString(), instanceMinorToString(), instanceRevisionToString(), instanceBuildToString() };

string[] args; string arg = args[0]; if(arg[0] == '-') { //This parameter is an option }





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However, { r < Q A b ^ l } r : = 2xr + b{r < 2xQ } . The increase in r, caused by the assignment r := 2xr + b, thus makes r at most Q more than the desired greatest value. Moreover, by (15.12) (see Exercise 15.17), {rmodQ = PmodQ }r := r-Q { r m o d Q = PmodQ } . Re-establishing r = P mod Q thus amounts to a conditional statement that subtracts Q from r in the case that r ^ Q. We have thus obtained the on-line remainder computation algorithm shown in Figure 15.6. Exercise 15.38. Suppose the input value b is not constrained to be a binary digit but satisfies 0 ^ b < B. (So the base is B.) Modify the algorithm accordingly. D

Not only can string characters be accessed individually using the array accessor, but it is also possible to retrieve the entire string as an array of characters using the string s ToCharArray() method Using this method, you could reverse the string using the SystemArrayReverse() method, as demonstrated in Listing 254, which determines whether a string is a palindrome

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class Palindrome { static void Main() { string reverse, palindrome; char[] temp; SystemConsoleWrite("Enter a palindrome: "); palindrome = SystemConsoleReadLine(); // Remove spaces and convert to lowercase reverse = palindromeReplace(" ", ""); reverse = reverseToLower(); // Convert to an array temp = reverseToCharArray(); // Reverse the array SystemArrayReverse(temp);

String dot = ""; writerWriteElementString("version", StringJoin(dot, parts)); } }

P,r := 0,0 ;

// Convert the array back to a string and // check if reverse string is the same if(reverse == new string(temp)) { SystemConsoleWriteLine("\"{0}\" is a palindrome", palindrome); } else { SystemConsoleWriteLine( "\"{0}\" is NOT a palindrome", palindrome); } } }

OUTPUT 222:

This method needs to be called from the CustomWriteElements() method, the remainder of which is copied verbatim from the DefaultWriteElements() method Complementary code must be written in the custom read methods to acquire the serialized string, split it into its constituents, and set the corresponding properties

{ Invariant: O ^ P A r = PmodQ } do true get.b { O^b^ 1 } ;

This example uses the new keyword; this time, it creates a new string from the reversed array of characters

Common Errors This section introduced the three different types of arrays: single-dimension, multidimensional, and jagged arrays Several rules and idiosyncrasies govern array declaration and use Table 27 points out some of the most common errors and helps solidify the rules Readers should consider reviewing the code in the Common Mistake column first (without looking at the Error Description and Corrected Code columns) as a way of verifying their understanding of arrays and their syntax

:= 2xP + fo , 2xr + b ;

int numbers[];

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