词法分析在Java中的字母

这个代码分析数字和运营商只我需要它来分析字母只，并把其他事情无效我应该做些什么细节我从教程网站得到这段代码，并修改它，但没有结果。词法分析在Java中的字母

 package lexical; 

    public class Tokenizer { 


int pos; 
char[] expression; 

Tokenizer(String expression) { 
    this.expression = expression.toCharArray(); 
    this.pos = 0; 
} 

enum Type { OPERATOR, LITTER, UNKNOWN } 

class Lexeme { 
    String type, token; 
    Lexeme(String type, String token) { 
     this.type = type; 
     this.token = token; 
    } 
} 

Lexeme getNextToken() { 
    StringBuilder token = new StringBuilder(); 
    boolean endOfToken = false; 
    Type type = Type.UNKNOWN; 
    while (!endOfToken && hasMoreTokens()) { 
     while(expression[pos] == ' ' && hasMoreTokens()) 
      pos++; 
     switch (expression[pos]) { 
      case '+': 
      case '-': 
      case '*': 
      case '/': 
       if(type != Type.LITTER) { 
        type = Type.OPERATOR; 
        token.append(expression[pos]); 
        pos++; 
       } 
       endOfToken = true; 
       break; 
      case ' ': 
       endOfToken = true; 
       pos++; 
       break; 
      default: 
       if(Character.isDigit(expression[pos]) || expression[pos] == '.') { 
        token.append(expression[pos]); 
        type = Type.LITTER; 
       } else { 
        System.out.println("Systax error at position: " + pos); 
       } 
       pos++; 
       break; 
     } 
    } 
    return new Lexeme(type.name().toLowerCase(), token.toString()); 
} 

boolean hasMoreTokens() { 
    return pos < expression.length; 
} 

public static void main(String[] args) { 
    String expression = "54+18+5"; 
    Tokenizer tokenizer = new Tokenizer(expression); 
    while (tokenizer.hasMoreTokens()) { 
     Lexeme nextToken = tokenizer.getNextToken(); 
     System.out.print("Type: " + nextToken.type + "\tLexeme: " + nextToken.token + "\n"); 
    } 
} 


}

来源

2014-12-06 user3513220

这里这部分是它正在寻找一个数字，给出了一个非数字或非周期性质的错误。

if(Character.isDigit(expression[pos]) || expression[pos] == '.') { 
    token.append(expression[pos]); 
    type = Type.LITTER; 
} else { 
    System.out.println("Systax error at position: " + pos); 
}

字符串可以normallly不放任何东西，所以你只想消除if语句，做

default: 
    token.append(expression[pos]);

但是，如果你想限制你的字符串中的字符，然后修改，如果条件检查你想要的角色。而不是：

if(Character.isDigit(expression[pos]) || expression[pos] == '.') {

将其更改为只接受您认为在字符串中有效的字符。例如

if (Character.isAlphabetic(expression[pos]) || expression[pos] == "-") {

您可能还需要一个新的标识，而不是

type = Type.LITTER;

来源

2014-12-06 13:15:21 Scooter

我努力，因为你发布你的答案，但没有你能不能请张贴检查所有字母和什么是所有代码无效在此先感谢 – user3513220 2014-12-06 14:02:15

什么字符是字母 - 只是AZ，az？ – Scooter 2014-12-06 14:18:25

字符是字母a-z – user3513220 2014-12-06 14:55:19

我不知道你想怎么处理“A12”或任何部分是字母和数字的一部分。此代码将返回“A”为“A12”

package lexical; 

public class Tokenizer { 

    int pos; 
    char[] expression; 

    Tokenizer(String expression) { 
     this.expression = expression.toCharArray(); 
     this.pos = 0; 
    } 

    enum Type { 

     OPERATOR, ALPHA, UNKNOWN, LITTER, ERROR 
    } 

    class Lexeme { 

     String type, token; 

     Lexeme(String type, String token) { 
      this.type = type; 
      this.token = token; 
     } 
    } 

    Lexeme getNextToken() { 
     StringBuilder token = new StringBuilder(); 
     boolean endOfToken = false; 
     Type type = Type.UNKNOWN; 
     // skip whitespace 
     while (expression[pos] == ' ' && hasMoreTokens()) { 
      pos++; 
     } 

     while (!endOfToken && hasMoreTokens()) { 

      switch (expression[pos]) { 
       case ' ': 
       { 
        endOfToken = true; 
        pos++; 
        break; 
       } 
       default: 
        if (Character.isAlphabetic(expression[pos])) {      
         token.append(expression[pos]); 
         type = Type.ALPHA;      
        } else { 
         System.out.println("Systax error at position: " + pos); 
        } 
        pos++; 
        break; 
      } 
     } 
     return new Lexeme(type.name().toLowerCase(), token.toString()); 
    } 

    boolean hasMoreTokens() { 
     return pos < expression.length; 
    } 

    public static void main(String[] args) { 
     expression = "Hello World"; 
     Tokenizer tokenizer = new Tokenizer(expression); 
     while (tokenizer.hasMoreTokens()) { 
      Lexeme nextToken = tokenizer.getNextToken(); 
      System.out.print("Type: " + nextToken.type + "\tLexeme: " + nextToken.token + "\n"); 
     }   
     expression = "123 ABC A12"; 
     tokenizer = new Tokenizer(expression); 
     while (tokenizer.hasMoreTokens()) { 
      Lexeme nextToken = tokenizer.getNextToken(); 
      System.out.print("Type: " + nextToken.type + "\tLexeme: " + nextToken.token + "\n"); 
     }   
    } 
}

输出：

Type: alpha Lexeme: Hello 
Type: alpha Lexeme: World 
Systax error at position: 0 
Systax error at position: 1 
Systax error at position: 2 
Type: unknown Lexeme: 
Type: alpha Lexeme: ABC 
Systax error at position: 9 
Systax error at position: 10 
Type: alpha Lexeme: A

来源

2014-12-06 16:09:28 Scooter

词法分析在Java中的字母

回答

相关问题