C#,ch9字符串和正则表达式.ppt

1、9 Strings and Regular Expressions,Shuai L 2012.9.5,0. Introduction,The string keyword in C# actually refers to the System.String .NET base class. System.Text System.Text.StringBuilder IFormatProvider, IFormattable System.Text.RegularExpressions,1. System.String,System.String is a class specifically

2、designed to store a string and allow a large number of operations on the string. You can concatenate strings using operator overloads string message1 = Hello; / returns Hello message1 += , There; / returns Hello, There string message2 = message1 + !; / returns Hello, There! C# also allows extraction

3、 of a particular character using an indexer-like syntax string message = Hello; char char4 = message4; / returns o. Note the string is zero-indexed,1. System.String,The key methods of System.String Compare, CompareOrdinal Concat, CopyTo, Insert, Join, Replace, Split, Substring Format IndexOf, IndexO

4、fAny, LastIndexOf, LastIndexOfAny PadLeft, PadRight ToLower, ToUpper, Trim,1. System.String,The String class has a shortcoming that makes it very inefficient for making repeated modifications to a given string it is actually an immutable data type, which means that after you initialize a string obje

5、ct, that string object can never change. The methods and operators that appear to modify the contents of a string actually create new strings, copying across the contents of the old string if necessary.,1. System.String,string greetingText = Hello from all the guys at Wrox Press. ; /39 chs greetingT

6、ext += We do hope you enjoy this book as much as we enjoyed writing it.; / 39+65-1=103 chs for(int i = z; i = a; i-) char old1 = (char)i; char new1 = (char)(i+1); greetingText = greetingText.Replace(old1, new1); for(int i = Z; i =A; i-) Console.WriteLine(Encoded:n + greetingText);,2. System.Text.Str

7、ingBuilder,The StringBuilder normally allocates more memory than is actually needed. Length which indicates the length of the string that it actually contains Capacity which indicates the maximum length of the string in the memory allocation Any modifications to the string take place within the bloc

8、k of memory assigned to the StringBuilder instance.,2. System.Text.StringBuilder,StringBuilder greetingBuilder = new StringBuilder(Hello from all the guys at Wrox Press. , 150); greetingBuilder.AppendFormat(We do hope you enjoy this book as much as we enjoyed writing it); for(int i = z; i=a; i-) cha

9、r old1 = (char)i; char new1 = (char)(i+1); greetingBuilder = greetingBuilder.Replace(old1, new1); for(int i = Z; i=A; i-) Console.WriteLine(Encoded:n + greetingBuilder);,2. System.Text.StringBuilder,The key methods of System.Text.StringBuilder Append, AppendFormat Insert, Remove, Replace ToString,3.

10、 Format Strings,Console.WriteLine() just passes the entire set of parameters to the static method, String.Format(). Console.WriteLine(The double is 0,10:E and the int contains 1, d, i); The implementation of the three-parameter overload of WriteLine() basically does this: public void WriteLine(strin

11、g format, object arg0, object arg1) this.WriteLine(string.Format(this.FormatProvider, format , new objectarg0, arg1); ,3. Format Strings,String.Format() now needs to construct the final string by replacing each format specifier with a suitable string representation of the corresponding object. Howev

12、er, as you saw earlier, for this process of building up a string you need a StringBuilder instance rather than a string instance.,3. Format Strings,In this example, a StringBuilder instance is created and initialized with the first known portion of the string, the text “The double is”. Next, the Str

13、ingBuilder.AppendFormat() method is called, passing in the first format specifier, 0,10:E, as well as the associated object, double, to add the string representation of this object to the string object being constructed. This process continues with StringBuilder.Append() and StringBuilder.AppendForm

14、at() being called repeatedly until the entire formatted string has been obtained.,3. Format Strings,StringBuilder.AppendFormat() has to figure out how to format the object. First, it probes the object to find out whether it implements an interface in the System namespace called IFormattable. You can

15、 determine this quite simply by trying to cast an object to this interface and seeing whether the cast succeeds, or by using the C# is keyword. If this test fails, AppendFormat() calls the objects ToString() method, which all objects either inherit from System.Object or override.,3. Format Strings,4

16、. IFormattable Interface,interface IFormattable string ToString(string format, IFormatProvider formatProvider); ,5. Vector,struct Vector: IFormattable public double x, y, z; public string ToString(string format, IFormatProvider formatProvider) if (format = null) return ToString(); string formatUpper

17、 = format.ToUpper(); switch (formatUpper) ,5. Vector,switch (formatUpper) case N:return | + Norm().ToString() + |; case VE:return String.Format( 0:E, 1:E, 2:E ), x, y, z); case IJK:StringBuilder sb = new StringBuilder(x.ToString(), 30); sb.AppendFormat( i + ); sb.AppendFormat(y.ToString(); sb.Append

18、Format( j + ); sb.AppendFormat(z.ToString(); sb.AppendFormat( k); return sb.ToString(); default:return ToString(); ,5. Vector,public override string ToString() return “( ” + x + “, ” + y + “, ” + z + “ )”; public double Norm() return x*x + y*y + z*z; static void Main() Vector v1 = new Vector(1,32,5); Vector v2 = new Vector(845.4, 54.3, -7.8); Console.WriteLine(nIn IJK format,nv1 is 0,30:IJKnv2 is 1,30:IJK, v1, v2); Console.WriteLine(nIn default format,nv1 is 0,30nv2 is 1,30, v1, v2); ,In IJK format, v1 is 1 i + 32 j + 5 k v2 is