15.2. Manipulating StringsThe String class provides a host of methods for comparing , searching, and manipulating strings , the most important of which are shown in Table 15-1. Table 15-1. String class properties and methods
15.2.1. Comparing StringsThe Compare( ) method of String is overloaded. The first version takes two strings and returns a negative number if the first string is alphabetically before the second, a positive number if the first string is alphabetically after the second, and zero if they are equal. The second version works just like the first but is case-insensitive. Example 15-1 illustrates the use of Compare( ) . Example 15-1. Compare( ) method
The output looks like this: compare s1: abcd, s2: ABCD, result: -1 Compare insensitive. result: 0 Example 15-1 begins by declaring two strings, s1 and s2 , and initializing them with string literals: string s1 = "abcd"; string s2 = "ABCD"; Compare( ) is used with many types. A negative return value indicates that the first parameter is less than the second, a positive result indicates the first parameter is greater than the second, and a zero indicates they are equal. In Unicode (as in ASCII), a lowercase letter has a smaller value than an uppercase letter; with strings identical except for case, lowercase comes first alphabetically. Thus, the output properly indicates that s1 (abcd) is "less than" s2 (ABCD): compare s1: abcd, s2: ABCD, result: -1 The second comparison uses an overloaded version of Compare( ) , which takes a third Boolean parameter, the value of which determines whether case should be ignored in the comparison. If the value of this "ignore case" parameter is true, the comparison is made without regard to case. This time the result is 0, indicating that the two strings are identical: Compare insensitive. result: 0 15.2.2. Concatenating StringsThere are a couple of ways to concatenate strings in C#. You can use the Concat( ) method, which is a static public method of the String class: string s3 = string.Concat(s1,s2); or you can simply use the overloaded concatenation ( + ) operator: string s4 = s1 + s2; Example 15-2 demonstrates both of these methods. Example 15-2. Concatenation
The output looks like this: s3 concatenated from s1 and s2: abcdABCD s4 concatenated from s1 + s2: abcdABCD In Example 15-2, the new string s3 is created by calling the static Concat( ) method and passing in s1 and s2, while the string s4 is created by using the overloaded concatenation operator ( + ) that concatenates two strings and returns a string as a result. 15.2.3. Copying StringsThere are two ways to copy strings. 99.9 percent of the time you will just write: oneString = theOtherString; and not worry about what is going on in memory. There is a second, somewhat awkward way to copy strings: myString = String.Copy(yourString); and this actually does something subtly different. The difference is somewhat advanced, but here it is in a nutshell . When you use the assignment operator ( = ), you create a second reference to the same object in memory, but when you use Copy , you create a reference to a new string that is initialized with the value of the first string. "Huh?" I hear you cry. An example will make it clear (see Example 15-3). Example 15-3. Copying strings
The output looks like this: string s1: abcd string s2 = s1; s1: abcd s2: abcd s1 == s2? True ReferenceEquals(s1,s2): True string s2 = string.Copy( s1 ); s1: abcd s3: abcd s1 == s3? True ReferenceEquals(s1,s3): False s1 = "Hello"; s1: Hello s2: abcd s1 == s2? False ReferenceEquals(s1,s2): False In Example 15-3, you start by initializing one string: string s1 = "abcd"; You then assign the value of s1 to s2 using the assignment operator: s2 = s1; You print their values, as shown in the first section of results, and find that not only do the two string references have the same value, as indicated by using the equality operator ( == ), but they actually point to the same object in memory, which is why ReferenceEquals returns true. On the other hand, if you create s3 and assign its value using String.Copy(s1) , while the two values are equal (as shown by using the equality operator), they refer to different objects in memory (as shown by the fact that ReferenceEquals returns false). Now, returning to s1 and s2 , which refer to the same object, if you change either one, for example, when you write: s1 = "Hello"; s3 goes on referring to the original string, but s1 now refers to a brand new string. If you later write: S3 = "Goodbye"; (not shown in the example), the original string referred to by s1 will no longer have any references to it, and it will be mercifully and painlessly destoryed by the Garbage Collector. 15.2.4. Testing for EqualityThe .NET String class provides three ways to test for the equality of two strings. First, you can use the overloaded Equals( ) method and ask one string (say, s6) directly whether another string (s5) is of equal value: Console.WriteLine( "\nDoes s6.Equals(s5)?: {0}", s6.Equals(s5)); You can also pass both strings to String 's static method Equals( ) : Console.WriteLine( "Does Equals(s6,s5)?: {0}" string.Equals(s6,s5)); Or you can use the String class's overloaded equality operator ( == ): Console.WriteLine( "Does s6==s5?: {0}", s6 == s5); In each of these cases, the returned result is a Boolean value (true for equal and false for unequal ). Example 15-4 demonstrates these techniques. Example 15-4. Are all strings created equal?
The output looks like this: s5 copied from s2: ABCD s6 = s5: ABCD Does s6.Equals(s5)?: True Does Equals(s6,s5)?: True Does s6==s5?: True The equality operator is the most natural of the three methods to use when you have two string objects. 15.2.5. Other Useful String MethodsThe String class includes a number of useful methods and properties for finding specific characters or substrings within a string, as well as for manipulating the contents of the string. Example 15-5 demonstrates a few methods, such as locating substrings, finding the index of a substring, and inserting text from one string into another. Following the output is a complete analysis. Example 15-5. Useful methods of the String class
The output looks like this: s5 copied from s2: ABCD String s3 is 4 characters long. The 5th character is r s3:Liberty Associates, Inc. provides custom .NET development, on-site Training and Consulting Ends with Training?: False Ends with Consulting?: True The first occurrence of Training in s3 is 73 s10: Liberty Associates, Inc. provides custom .NET development, on-site excellent Training and Consulting s11: Liberty Associates, Inc. provides custom .NET development, on-site excellent Training and Consulting The Length property returns the length of the entire string, and the index operator ( [] ) is used to access a particular character within a string: Console.WriteLine( "\nString s3 is {0} characters long. ", s5.Length); Console.WriteLine( "The 5th character is {0}\n", s3[4]); Here's the output: String s3 is 4 characters long. The 5th character is r The EndsWith( ) method asks a string whether a substring is found at the end of the string. Thus, you might first ask if s3 ends with "Training" (which it does not), and then if it ends with "Consulting" (which it does): Console.WriteLine("s3:{0}\nEnds with Training?: {1}\n", s3, s3.EndsWith("Training") ); Console.WriteLine( "Ends with Consulting?: {0}", s3.EndsWith("Consulting")); The output reflects that the first test fails and the second succeeds: Ends with Training?: False Ends with Consulting?: True The IndexOf( ) method locates a substring within a string, and the Insert( ) method inserts a new substring into a copy of the original string. The following code locates the first occurrence of "Training" in s3 : Console.WriteLine("\nThe first occurrence of Training "); Console.WriteLine ("in s3 is {0}\n", s3.IndexOf("Training")); The output indicates that the offset is 73: The first occurrence of Training in s3 is 73 Then use that value to insert the word "excellent," followed by a space, into that string. Actually the insertion is into a copy of the string returned by the Insert( ) method and assigned to s10 : string s10 = s3.Insert(73,"excellent "); Console.WriteLine("s10: {0}\n",s10); Here's the output: s10: Liberty Associates, Inc. provides custom .NET development, on-site excellent Training and Consulting Finally, you can combine these operations to make a more efficient insertion statement: string s11 = s3.Insert(s3.IndexOf("Training"),"excellent "); Console.WriteLine("s11: {0}\n",s11); with the identical result: s11: Liberty Associates, Inc. provides custom .NET development, on-site excellent Training and Consulting 15.2.6. Finding SubstringsThe String class has methods for finding and extracting substrings . For example, the IndexOf( ) method returns the index of the first occurrence of a string (or of any character in an array of characters) within a target string. For example, given the definition of the string s1 as: string s1 = "One Two Three Four"; you can find the first instance of the characters "hre" by writing: int index = s1.IndexOf("hre"); This code sets the int variable index to 9, which is the offset of the letters "hre" in the string s1 . Similarly, the LastIndexOf( ) method returns the index of the last occurrence of a string or substring. While the following code: s1.IndexOf("o"); returns the value 6 (the first occurrence of the lowercase letter "o" is at the end of the word "Two"), the method call: s1.LastIndexOf("o"); returns the value 15 (the last occurrence of "o" is in the word "Four"). The Substring( ) method returns a 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 you ( optionally ) provide. Example 15-6 illustrates the Substring( ) method. Example 15-6. Finding substrings by index
The output looks like this: s2: Four s3: Three s4: Two s5: One s1: One Example 15-6 is not the most elegant solution possible to the problem of extracting words from a string, but it is a good first approximation , and it illustrates a useful technique. The example begins by creating a string, s1 : string s1 = "One Two Three Four"; The local variable index is assigned the value of the last literal space in the string (which comes before the word "Four"): index=s1.LastIndexOf(" "); The substring that begins one position later is assigned to the new string, s2 : string s2 = s1.Substring(index+1); This extracts the characters from index +1 to the end of the line (the string "Four") and assigns the value "Four" to s2 . The next step is to remove the word "Four" from s1 ; assign to s1 the substring of s1 that begins at 0 and ends at the index: s1 = s1.Substring(0,index);
You reassign index to the last (remaining) space, which points you to the beginning of the word "Three." You then extract the character "Three" into string s3 . Continue like this until you've populated s4 and s5 . Finally, display the results: s2: Four s3: Three s4: Two s5: One s1: One 15.2.7. Splitting StringsA more effective solution to the problem illustrated in Example 15-6 would be to use the String class's Split( ) method, which parses a string into substrings. To use Split( ) , pass in an array of delimiters (characters that indicate where to divide the words). The method returns an array of substrings (which Example 15-7 illustrates). The complete analysis follows the code. Example 15-7. The Split( ) method
The output looks like this: 1: One 2: Two 3: Three 4: Liberty 5: Associates 6: 7: Inc. Example 15-7 starts by creating a string to parse: string s1 = "One,Two,Three Liberty Associates, Inc."; The delimiters are set to the space and comma characters. Then call Split( ) on the string, passing in the delimiters: String[] resultArray = s1.Split(delimiters); Split( ) returns an array of the substrings that you can then iterate over using the foreach loop, as explained in Chapter 10: foreach (String subString in resultArray)
Start the foreach loop by initializing output to an empty string, and then build up the output string in four steps. Start by concatenating the incremented value of ctr to the output string, using the += operator. output += ctr++; Next add the colon , then the substring returned by Split( ) , and then the newline: output += ": "; output += subString; output += "\n"; With each concatenation, a new copy of the string is made, and all four steps are repeated for each substring found by Split( ) . This repeated copying of 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 . 15.2.8. The StringBuilder ClassYou 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
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. Example 15-8. The StringBuilder class
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: One 2: Two 3: Three 4: Liberty 5: Associates 6: 7: 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 previous 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. |