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1: 2: 3: 4: 5: One Two Three Liberty Associates
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Manipulating Strings |
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Example 15-7 starts by creating a string to parse:
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string s1 = "One,Two,Three Liberty Associates, Inc.";
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The delimiters are set to the space and comma characters. Then call Split( ) on the string, passing in the delimiters:
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String[] resultArray = s1.Split(delimiters);
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Split( ) returns an array of the substrings that you can then iterate over using the foreach loop, as explained in 10:
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foreach (String subString in resultArray)
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You can, of course, combine the call to split with the iteration, as in the following:
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foreach (string subString in s1.Split(delimiters))
C# programmers are fond of combining statements like this. The advantage of splitting the statement into two, however, and of using an interim variable like resultArray is that you can examine the contents of resultArray in the debugger.
Start the foreach loop by initializing output to an empty string, and then create each line of the output in three steps. Start with the incremented value ctr. Then use the += operator to add the colon, then the substring returned by Split( ):
Console.WriteLine(ctr++ + ":" + subString);
With each concatenation, a new copy of the string is made, and all three steps are repeated for each substring found by Split( ). This repeated copying of the string is terribly inefficient. The problem is that the string type is not designed for this kind of operation. What you want is to create a new string by appending a formatted string each time through the loop. The class you need is StringBuilder.
The StringBuilder Class
You can use the System.Text.StringBuilder class for creating and modifying strings. Table 15-2 summarizes the important members of StringBuilder.
Table 15-2. StringBuilder members Method or property
Append( ) AppendFormat( )
Explanation Overloaded public method that appends a typed object to the end of the current
StringBuilder
Overloaded public method that replaces format specifiers with the formatted value of an object
|
15: Strings
Table 15-2. StringBuilder members (continued) Method or property
EnsureCapacity( ) Capacity Insert( ) Length MaxCapacity Remove( ) Replace( )
Explanation Ensures that the current StringBuilder has a capacity at least as large as the specified value Property that retrieves or assigns the number of characters the StringBuilder is capable of holding Overloaded public method that inserts an object at the specified position Property that retrieves or assigns the length of the StringBuilder Property that retrieves the maximum capacity of the StringBuilder Removes the specified range of characters Overloaded public method that replaces all instances of the specified characters with new characters
Unlike String, StringBuilder is mutable; when you modify an instance of the StringBuilder class, you modify the actual string, not a copy. Example 15-8 replaces the String object in Example 15-7 with a StringBuilder object.
using using using using System; System.Collections.Generic; System.Linq; System.Text;
namespace Example_15_8_ _ _ _StringBuilder { class Tester { public void Run( ) { // create some strings to work with string s1 = "One,Two,Three Liberty Associates, Inc."; // constants for the space and comma characters const char Space = ' '; const char Comma = ','; // array of delimiters to split the sentence with char[] delimiters = new char[] { Space, Comma }; // use a StringBuilder class to build the // output string
Manipulating Strings |
StringBuilder output = new StringBuilder( ); int ctr = 1; // split the string and then iterate over the // resulting array of strings foreach (string subString in s1.Split(delimiters)) { // AppendFormat appends a formatted string output.AppendFormat("{0}: {1}\n", ctr++, subString); } Console.WriteLine(output); } static void Main( ) { Tester t = new Tester( ); t.Run( ); } } }
Only the last part of the program is modified. Rather than using the concatenation operator to modify the string, use the AppendFormat( ) method of StringBuilder to append new formatted strings as you create them. This is much easier and far more efficient. The output is identical:
1: 2: 3: 4: 5: 6: 7: One Two Three Liberty Associates Inc.
Because you passed in delimiters of both comma and space, the space after the comma between Associates and Inc. is returned as a word, numbered 6 in the preceding code. That is not what you want. To eliminate this, you need to tell Split( ) to match a comma (as between One, Two, and Three), a space (as between Liberty and Associates), or a comma followed by a space. It is that last bit that is tricky and requires that you use a regular expression.
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