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Java 如何使用StringTokenizer类_Java_Stringtokenizer - Fatal编程技术网

Java 如何使用StringTokenizer类

Java 如何使用StringTokenizer类,java,stringtokenizer,Java,Stringtokenizer,我应该通过使用StringTokenizer类将字符串拆分为标记来计算字符串。之后,我应该使用“Integer.parseInt”将这些标记转换为int值 我不明白的是,在分割代币之后,我应该如何处理它们 public class Tester { public static void main(String[] args) { String i = ("2+5"); StringTokenizer st = new StringTokenizer(i, "+-", t

我应该通过使用StringTokenizer类将字符串拆分为标记来计算字符串。之后,我应该使用“Integer.parseInt”将这些标记转换为int值

我不明白的是,在分割代币之后,我应该如何处理它们

public class Tester {

    public static void main(String[] args) {
    String i = ("2+5");
    StringTokenizer st = new StringTokenizer(i, "+-", true);
    while (st.hasMoreTokens()) {
        System.out.println(st.nextToken());
    }
    int x = Integer.parseInt();
//what exactly do I have to type in here, do convert the token(s) to an int value?
}

}
如果我现在明白了,我现在有三个代币。这将是:“2”,“+”和“5”

如何准确地将这些标记转换为int值? 我必须分别转换它们吗


非常感谢您的帮助。

也许您可以使用:

    String i = ("2+5");
    StringTokenizer st = new StringTokenizer(i, "+-", true);
    while (st.hasMoreTokens()) {
        String tok=st.nextToken();
        System.out.println(tok);

        //what exactly do I have to type in here, do convert the token(s) to an int value?
        if ("+-".contains(tok)) {
            //tok is an operand
        }
        else {
            int x = Integer.parseInt(tok);
        }       
    }

也许你可以用这个:

    String i = ("2+5");
    StringTokenizer st = new StringTokenizer(i, "+-", true);
    while (st.hasMoreTokens()) {
        String tok=st.nextToken();
        System.out.println(tok);

        //what exactly do I have to type in here, do convert the token(s) to an int value?
        if ("+-".contains(tok)) {
            //tok is an operand
        }
        else {
            int x = Integer.parseInt(tok);
        }       
    }

要使用从字符串中提取的整数进行一些计算,必须将它们放入ArrayList中。您必须使用try/catch操作来避免NumberFormatException。此外,您还可以直接从ArrayList中获取值,并根据需要使用它们。例如:

public static void main(String[] args) {
    ArrayList <Integer> myArray = new ArrayList <>();    
    String i = ("2+5");
        StringTokenizer st = new StringTokenizer(i, "+-/*=", true);
        while (st.hasMoreTokens()) {
            try {
            Integer stg = Integer.parseInt(st.nextToken(i));
            myArray.add(stg);
                }
            catch (NumberFormatException nfe) {};
            }

       System.out.println("This is an array of Integers: " + myArray);
       for (int a : myArray) {
           int x = a;
           System.out.println("This is an Integer: " + x);
       }
       int b = myArray.get(0);
       int c = myArray.get(1);
       System.out.println("This is b: " + b);
       System.out.println("This is c: " + c);
       System.out.println("This is a sum of b + c: " + (b + c));

}

要使用从字符串中提取的整数进行一些计算,必须将它们放入ArrayList中。您必须使用try/catch操作来避免NumberFormatException。此外,您还可以直接从ArrayList中获取值,并根据需要使用它们。例如:

public static void main(String[] args) {
    ArrayList <Integer> myArray = new ArrayList <>();    
    String i = ("2+5");
        StringTokenizer st = new StringTokenizer(i, "+-/*=", true);
        while (st.hasMoreTokens()) {
            try {
            Integer stg = Integer.parseInt(st.nextToken(i));
            myArray.add(stg);
                }
            catch (NumberFormatException nfe) {};
            }

       System.out.println("This is an array of Integers: " + myArray);
       for (int a : myArray) {
           int x = a;
           System.out.println("This is an Integer: " + x);
       }
       int b = myArray.get(0);
       int c = myArray.get(1);
       System.out.println("This is b: " + b);
       System.out.println("This is c: " + c);
       System.out.println("This is a sum of b + c: " + (b + c));

}

必须将从
st.nextToken()
返回的值存储在字符串变量中。而且,文档是你的朋友;应该对您有所帮助。现在您正在将令牌打印到标准输出,但没有将其保存到任何地方。考虑在数组中保存圣NEXTROKEN()的输出,然后您可以稍后访问它来执行整数。PSESETIN(),您必须在字符串变量中存储从<代码>圣NEXTROKEN()/<代码>返回的值。而且,文档是你的朋友;应该对您有所帮助。现在您正在将令牌打印到标准输出,但没有将其保存到任何地方。考虑在数组中保存圣NeXToTK()的输出,然后您可以稍后访问它来执行整数。