Question

这是我的代码块。我正在使用for循环遍历字符串，拉出令牌（一个用空格分隔的单词），确定它是否是保留字，如果这样，它将显示“ Reserved word is：”，否则将显示“当前单词是：“。到目前为止，这是我所得到的，我仍然坚持如何使isKeyWord工作。注意：我不能使用分割功能，扫描器或令牌生成器。只是一个for循环。

public class Week3Project {
    final static String program = "/*\n" +
" * To change this license header, choose License Headers in Project Properties.\n" +
" * To change this template file, choose Tools | Templates\n" +
" * and open the template in the editor.\n" +
" */\n" +
"package testapplication2;\n" +
"\n" +
"import java.util.Scanner;\n" +
"\n" +
"/**\n" +
" *\n" +
" * @author james\n" +
" */\n" +
"public class TestApplication2 {\n" +
"\n" +
"    /**\n" +
"     * @param args the command line arguments\n" +
"     */\n" +
"    public static void main(String[] args) {\n" +
"        Scanner input = new Scanner(System.in);\n" +
"        \n" +
"        System.out.println(\"Enter integer #1\");\n" +
"        int num1 = input.nextInt();\n" +
"        \n" +
"        System.out.println(\"Enter integer #2\");\n" +
"        int num2 = input.nextInt();\n" +
"        \n" +
"        System.out.println(\"Enter integer #3\");\n" +
"        int num3 = input.nextInt();\n" +
"        \n" +
"        System.out.println(\"Enter integer #4\");\n" +
"        int num4 = input.nextInt();\n" +
"        \n" +
"        System.out.println(\"Enter integer #5\");\n" +
"        int num5 = input.nextInt();\n" +
"        \n" +
"        //determine the sum\n" +
"        int sum = num1 + num2 + num3 + num4 + num5;\n" +
"        \n" +
"        //this is helpful to make sure your sum is correct\n" +
"        System.out.println(\"The sum is: \" + sum);\n" +
"        \n" +
"        //why doesn't this generate the sum correctly\n" +
"        double average = sum / 5;\n" +
"        \n" +
"        //The average, lets hope its right...\n" +
"        System.out.println(\"The average of your numbers is: \" + average);\n" +
"        \n" +
"    }\n" +
"    \n" +
"}\n" +
"";


    public static void main(String[] args)
    {
        String str = program;

        String s = "";
        String[] keywords = {"abstract", "assert", "boolean", "break", "byte", "case", "catch", "char", "class", "const", "continue", "default", "do", "double", "else", "enum",
 "extends", "final", "finally", "float", "for", "goto", "if", "implements", "import", "instanceof", "int", "interface", "long", "native", "new", "package", "private", "protected", "public", "return", "short", "static",
 "strictfp", "super", "switch", "synchronized", "this", "throw", "throws", "transient", "try", "void", "volatile", "while"};
        for (int i = 0; i < str.length(); i++) {
            s += str.charAt(i) + "";
            if (str.charAt(i) == ' ' || str.charAt(i) == '\t' || str.charAt(i) == '\n') {
                String currentWord = s.toString();
                //System.out.println(currentWord);                
                boolean isKeyWord = false; 
                for (int j = 0; j < keywords.length; j++) { 
                    if (currentWord.equalsIgnoreCase(keywords[j])) { 
                        isKeyWord = true;
                    }
                } break; } }
                if (isKeyWord == true) {
                    System.out.println("Reserved word is: ["  + currentWord + "]");
                }
                else {
                    System.out.println("Current word is: [" + currentWord + "]")
                }
                s = "";//Clear the string to get it ready to build next token.


            }
        }

Answer 1

    for (int i = 0; i < str.length(); i++) {
        if (str.charAt(i) == ' ' || str.charAt(i) == '\t' || str.charAt(i) == '\n') {
            // end of token, check if key word
            String currentWord = s.toString();     
            boolean isKeyword = false       
            for (int j = 0; j < keywords.length; j++) { 
                if (currentWord.equalsIgnoreCase(keywords[j])) { 
                    isKeyword = true;
                    break;
                } 
            }
            if(isKeyword) {
                System.out.println("Reserved word is: ["  + currentWord + "]");
            } else {
                System.out.println("Current word is: [" + currentWord + "]");
            }
            s = "";//Clear the string to get it ready to build next token.
        } else {
            // continue building token
            s += str.charAt(i) + "";
        }
    }

您的循环版本会在检查之前添加空格/制表符/换行符，这会导致当前单词与您的关键字永远不匹配。

Answer 2

首先，我可能会将包含String[]的{{1}}移到keywords字段中（但是我没有在这里粘贴它，因为它很长-而且不会< strong> 需要进行更改）。接下来，在所有这些串联中首选static final。 StringBuilder是确定您是否遇到空白的好方法。像

Character.isWhitespace(char)

而且，在Java 8+中，数组搜索可能可以通过StringBuilder sb = new StringBuilder(); for (int i = 0; i < program.length(); i++) { char ch = program.charAt(i); if (!Character.isWhitespace(ch)) { sb.append(ch); } else { String token = sb.toString(); if (token.isEmpty()) { continue; } boolean isKeyWord = false; for (String word : keywords) { if (word.equals(token)) { isKeyWord = true; break; } } System.out.printf("%s word is : [%s]%n", isKeyWord ? "Reserved" : "Current", token); sb.setLength(0); } }和Arrays.stream来完成

anyMatch

Answer 3

检查是否可以帮助您

public static void main(String[] args) {
String[] keywords = {"abstract", "assert", "boolean", "break", "byte", "case", "catch", "char", "class", "const", "continue", "default", "do", "double", "else", "enum",
    "extends", "final", "finally", "float", "for", "goto", "if", "implements", "import", "instanceof", "int", "interface", "long", "native", "new", "package", "private",
    "protected", "public", "return", "short", "static",
    "strictfp", "super", "switch", "synchronized", "this", "throw", "throws", "transient", "try", "void", "volatile", "while"};

 for (String value : keywords) {
   if (program.contains(value)) {
     System.out.println("Reserved word is: [" + value + "]");
   }
 }
}

我的结果

保留字是：[class]
保留字是：[do]
保留字是：[double]
保留字是：[import]
保留字是：[int]
保留字为：[新]
保留字是：[package]
保留字是：[public]
保留字是：[静态]
保留字是：[this]
保留字是：[void]

我的代码中缺少什么？爪哇

3 个答案: