Question

我正在尝试编写一个字母频率程序，用于计算.txt文件中的字母字符，并将频率数据显示在2列表中。我正在努力从函数processFile中调用函数char2int。我不确定是否需要使用if语句。另外，我总结了所有字母字符和其他字符。如果有人可以帮助解释我做错了什么，我将不胜感激。

这是我的代码：

final static int AlphabetSize = 26;
final static Scanner cin = new Scanner(System.in);
final static PrintStream cout = System.out;
final static int MaxBarLength = 50;

public static void main(String[] args) {
    String fileName;


    // get the file name
    cout.print("Enter the file name: ");
    fileName = cin.nextLine();

    // process the file
    try {
        processFile(fileName);
    }
    catch (IOException e) {
        e.printStackTrace();
    } // end try


} // end main

static void processFile(final String fileName) 
        throws FileNotFoundException, IOException 
{ 
    FileInputStream inFile = new FileInputStream(fileName);
    int inputValue;

    // declare other variables you need
            int counters [] = new int [26];
               int alpha = 0;
               int num = 0;
               int ticker = 0;

    // get the first character from file
    inputValue = inFile.read();
    while (inputValue != -1) {
        char ch = (char) inputValue;

        // add code to process this character
                  if (){
            counters[char2int(ch)]++;
                    alpha++;
                }else{
                      num++;
                  }
        // read next input character
        inputValue = inFile.read();
    } // end loop

    inFile.close();
        System.out.println("\nThe data file has " + alpha + " alphabetic, and " + num + " other characters.\n");
    // generate appropriate output
            display(counters);


} // end function

static void display(final int [] counters) {
    // write code for this function

    System.out.println("Letter" + " " + "Count" + " " + "Bar");
    System.out.println("------" + " " + "-----" + " " + "---");
   printChars(counters);
   } // end function

// char2int is complete
static int char2int(final char arg) {
    if (!Character.isLetter(arg))
        return -1;
    else
        return (int) Character.toUpperCase(arg) - (int) 'A';
} // end function


// function printChars writes n copies of the character c to the
// standard output device
static void printChars (final int[] counters) {
    // write the code
    for (int i = 0; i < 26; i++){
       System.out.printf("%c%7d\n", i + 'A', counters[i]);

   }


 // end printChars


} // end function

Answer 1

我个人会这样做威廉：

int c = char2int(ch);
if (c >= 0) { 
    counters[c]++;
    alpha++;
}
else { num++; }

如果文本文件中有任何字符，您还应该考虑使用非字母字符。这也可以在 char2int（）方法中处理。

Letter Frequency程序函数调用Java

1 个答案: