Question

我是Java的新手，并且练习我尝试制作一个使用某些值（主要是字母）生成新的随机单词的程序。该程序旨在从文本文件中获取这些值。下一步是定义包含其类型的每个字母的清单（Arrays）（首先通过为每个int的长度定义变量（array），然后填充每个数组用正确的字母（String））。在检查我的进度时，我意识到我的代码并没有更新库存长度（cInv和vInv）。

这是代码的相关部分：

static File language;
static Scanner scanFile;
static Scanner scanInput = new Scanner(System.in);

static int cInv;
static int vInv;

//Getters go here

public static void setLanguage(File language) {Problem.language = language;}
public static void setCInv(int CInv) {Problem.cInv = cInv;}
public static void setVInv(int VInv) {Problem.vInv = vInv;}

//Asks for the file with the language values.
public static void takeFile() throws FileNotFoundException {
    String route = scanInput.nextLine();
    setLanguage(new File(route));

    BufferedReader br;
    br = new BufferedReader(new FileReader(language));
}

//Gathers the language values from the file.
public static void readFile() throws FileNotFoundException {
    takeFile();
    scanFile = new Scanner(language);

    //Defines the inventory sizes. It seems the failure is here.
    if (scanFile.hasNextInt()) {setCInv(scanFile.nextInt());}
    if (scanFile.hasNextInt()) {setVInv(scanFile.nextInt());}
}

public static void main(String[] args) throws FileNotFoundException {
    readFile();
    //The following line is for checking the progress of my coding.
    System.out.println(cInv);
}

这是它读取（或应该阅读）的文本文件的相关部分：

---Phonemes---  
Consonants: 43  
Vowels:     9

它的输出为0

我已尝试在文件的最开头键入43，我也尝试在输入中输入数字，但我仍然保持0。有人知道我错过了什么或做错了吗？

Answer 1

首先，在重新分配相同的静态变量时更改分配。

public static void setCInv(int CInv) {Problem.cInv = CInv;}
public static void setVInv(int VInv) {Problem.vInv = CInv;}

第二，您需要移动文件中的所有标记以识别数字并更新相应的变量。

//Gathers the language values from the file.
public static void readFile() throws FileNotFoundException {
    takeFile();
    scanFile = new Scanner(language);
    int num = 0;
    scanFile.nextLine(); //Skip ---Phonemes---
    setCInv(getInt(scanFile.nextLine()));
    setVInv(getInt(scanFile.nextLine()));
}

public static int getInt(String str){
    System.out.println(str);
    int num =0;
    Scanner line = new Scanner(str);
    //Splits the scanned line into tokens (accessed via next()) and search for numbers.
    //Similar thing could have been done using String.split(token);
    while(line.hasNext()){
        try{
            num = Integer.parseInt(line.next());
            return num;
        }catch(NumberFormatException e){}
    }
    return num;
}

scanner.hasNextInt（）返回意外值

1 个答案: