Ivon神兜兜

暑假用Java写的一段很简单的识别浮点数的代码。不想翻译了就直接放上来，大家对照状态图看源程序便知道是怎么回事了。

EBNF:

<floating-point)->{(0|1|2|3|4|5|6|7|8|9)}+.{(0|1|2|3|4|5|6|7|8|9)}+[e(+|-){(0|1|2|3|4|5|6|7|8|9)}+]

Definition of floating-point literals in Java:
Java对于浮点数的定义如下：

A floating point literal have the following parts: a decimal integer, a decimal point, a fraction, an exponent (starts with an “e” or E followed by an integer, which could be preceded by “+” or “-”).

A floating-point literal can have the following parts: a decimal integer, a decimal point (“.”), a fraction (another decimal number), an exponent, and a type suffix. The exponent part is an “e” or “E” followed by an integer, which can be signed (preceded by “+” or “-”). A floating-point literal must have at least one digit, plus either a decimal point or “e” (or “E”).

According to the definition, the state diagram of floating-point parser is:
有穷状态机图如下：

Click to view the source code.
点击查看源代码。 
package fp;
import java.io.Console;
/**
 *
 * @author Ivon
 */
public class Main {
    char nextChar;
    public static void main(String[] args) {
        int x=0, y=0, z=0;
        /*x is used to record the position of the decimal point
         * y is used to recored the position of "e" and "E"
         */
        Console c = System.console();
        if (c == null) {
            System.err.println("No console.");
            System.exit(1);
        }
        String thenumber = c.readLine("Enter a number: ");
        if (isDigit(thenumber,0)== -1){
            System.out.println(thenumber + " is an integer.");

        }else
            //The next step is to judge if the input is only a number with decimal point
            {
            x=isDigit(thenumber,0);
            if (thenumber.charAt(x)== '.'){
                x=isNumberWithPoint(thenumber,x);
                if (x>=0) {
                    if (thenumber.charAt(x)=='e'||thenumber.charAt(x)=='E'){
                        isValidExp(thenumber,x);
                    }else{
                        System.out.println(thenumber + " is not a valid float point number.");
                    }
                }
            }else if(thenumber.charAt(x)== 'E'||thenumber.charAt(x)== 'e'){
                isValidExp(thenumber, x);
            }else{
                System.out.println(thenumber + " is not a valid float point number.");
            }

        }

    }

    public static int isDigit(String s, int number){
        int isDigitFlag=-999;
        for (int i=number; i<s.length();i++){
            if ( !Character.isDigit(s.charAt(i)) ) {
                isDigitFlag = i;
                break;
            }else{
                isDigitFlag = -1;
            }
        }
        return (isDigitFlag);

    }

    public static int isNumberWithPoint(String thenumber,int x){
                if (x+1==thenumber.length()){
                    System.out.println(thenumber+" is a number with decimal point.");
                    return -1;
                }
                if (isDigit(thenumber, x+1) == -1){
                    System.out.println(thenumber+" is a number with decimal point.");
                    return -1;
                }else{
                    return isDigit(thenumber, x+1); //return the position of the next non-digit
                }

    }

    public static int isValidExp(String thenumber, int x){
        if (x+1==thenumber.length()){
            System.out.println(thenumber+" is a valid float point number.");
            return 1;

        }
                if(thenumber.charAt(x+1)=='+'||thenumber.charAt(x+1)=='-'){
                    x++;
                }

                    if (isDigit(thenumber, x+1) == -1){
                    System.out.println(thenumber+" is a valid float point number.");
                    }else{
                    System.out.println(thenumber+" is not a valid float point number.");
                    }
         return 1;
    }

}

Test Case:

 以下是测试用例: 
Integer:
11 – Pass
hello - Fail
$$$ - fail
Number with float point:
11.11 – Pass
11..11 – fail
1abc1 – fail
Float Point Number:
11e – Pass
11w – fail
11e11 – pass
11ee11 – fail
11e-1 – pass
11e+1 – pass
11e++1 – pass
11e+1w – fail
Should you have any better approach, please don't hesitate to correct me.

 如果有误，欢迎指正