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datamatrix net example

datamatrix net examples

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returns the value 6 (the first occurrence of the lowercase letter o is at the end of the word Two), the method call:

Deletion of a record, done by setting a tombstone into the record, is also done using this process Equation 3-43 is always appropriate for systems which disallow both the updating of key elds and variable-length records In the general case, the previous version of the record has to be deleted and the new record inserted appropriately The old record is rewritten with the tombstone; the key and pointer elds are kept intact so that the structure of the le is not violated Then TU = TF + TRW + TI = 2TF + 7r + btt in general 3-44 If the cases which permit in-place updates are recognized by the le system, then TU is to be computed based on the mix of in-place and general updates

s1.LastIndexOf("o");

Sec 3-3

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returns the value 15 (the last occurrence of o is in the word Four). The Substring( ) method returns a contiguous series of characters. You can ask it for all the characters starting at a particular offset and ending either with the end of the string or with an offset that you (optionally) provide. Example 15-6 illustrates the Substring( ) method.

An exhaustive search of the le has to be made when the search argument is not the indexed attribute The le may be read serially by following the over ow chains for every block, or if seriality is not required, the entire data area on a cylinder can be read sequentially, followed by sequential reading of the entire over ow area In either case, the index can be ignored unless it contains space-allocation information Most systems provide only the ability to read serially, so that

using using using using System; System.Collections.Generic; System.Linq; System.Text;

serial 3-45

namespace Example_15_6_ _ _ _Finding_Substrings { class Tester { public void Run( ) { // create some strings to work with string s1 = "One Two Three Four"; int index; // get the index of the last space index = s1.LastIndexOf(" ");

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The assumptions leading to Eq 3-41 are valid here In the alternative case the evaluation would consider the e ective transfer rate, neglecting the delay when skipping from data blocks to over ow blocks Unused over ow blocks will not be read, so that o can be used to estimate the size of the over ow areas read Now TX (n + o ) R t sequential 3-46

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To reorganize the old le, the entire le is read serially and rewritten without the use of over ow areas As a by-product a new index is constructed The prior index can be ignored, since the le is read serially Additional bu ers in memory are needed to collect the new data and index information For the index it is desirable to have at least one bu er for every level The new main le should be at least double-bu ered All outputs can be written sequentially Then

// get the last word. string s2 = s1.Substring(index + 1); // // // s1 set s1 to the substring starting at 0 and ending at index (the start of the last word) thus s1 has "one two three" = s1.Substring(0, index);

3-47

// find the last space in s1 (after two) index = s1.LastIndexOf(" "); // set s3 to the substring starting at // index, the space after "two" plus one more // thus s3 = "three" string s3 = s1.Substring(index + 1); // reset s1 to the substring starting at 0 // and ending at index, thus the string "one two" s1 = s1.Substring(0, index); // reset index to the space between // "one" and "two" index = s1.LastIndexOf(" "); // set s4 to the substring starting one // space after index, thus the substring "two" string s4 = s1.Substring(index + 1); // reset s1 to the substring starting at 0 // and ending at index, thus "one" s1 = s1.Substring(0, index); // set index to the last space, but there is // none so index now = -1 index = s1.LastIndexOf(" "); // set s5 to the substring at one past // the last space. there was no last space // so this sets s5 to the substring starting // at zero string s5 = s1.Substring(index + 1); Console.WriteLine("s2: {0}\ns3: {1}", s2, s3); Console.WriteLine("s4: {0}\ns5: {1}\n", s4, s5); Console.WriteLine("s1: {0}\n", s1); } static void Main( ) { Tester t = new Tester( ); t.Run( ); } } }

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