c++ - 字长频率，程序只打开cmd，什么都不做，C++

Question

我一直在做一项任务。我已经尝试了很多解决方案，但我无法找到一种方法来使其工作。我的任务是创建一个代码，从文件中读取文本并显示单词长度的频率。也就是说，如果它读取“My name is Jon”，它应该显示“1 = 0, 2 = 2, 3 = 1, 4 = 1”(第一个数字是单词的长度，第二个是频率)。我写了一段代码，我很确定它可以正常工作，但它不起作用，它甚至不显示错误或什么都不显示，它只是打开 cmd 而什么都不做。这是我的代码:

#include <iostream>
#include <ctype.h>
#include <iomanip>
#include <fstream>

using namespace std;

int NextWordLength(void); //function prototypes
void DisplayFrequencyTable(const int Words[]);

const int WORD_LENGTH = 16; // global constant for array

int main()
{
    int WordLength;  //actual length of word 0 to x
    int NumOfWords[WORD_LENGTH] = {0}; //array hold # of lengths of words

    WordLength=NextWordLength();
    while (WordLength)  //continue to loop until no word i.e. 0
    {
        (WordLength <= 14) ? (++NumOfWords[WordLength]):(++NumOfWords[15]);
        WordLength=NextWordLength();
    }
    DisplayFrequencyTable(NumOfWords);
}

int NextWordLength(void)
{
    fstream fin ("in.txt", ios::in);
    char Ch;
    int EndOfWord = 0; //tells when we have read in one word
    int LengthOfWord = 0;

    Ch = cin.get();  //get first character

    while (!cin.eof() && !EndOfWord)
    {
        while (isspace(Ch) || ispunct(Ch)) //skips elading white spaces
        {
            Ch = cin.get(); //and leading punctation marks
        }
        if (isalnum(Ch)) // if character is a letter or number
        {
            ++LengthOfWord;
        }
        Ch = cin.get(); //get next character
        if((Ch=='-')&&(cin.peek()=='\n')) //check for hyphenated word over two lines
        {
            Ch = cin.get();
            Ch = cin.get();

        }
        if ((Ch=='-')&&(isalpha(cin.peek()))) // check for hyphenated word in one line
        {
            ++LengthOfWord; //count the hyphen as part of word
            Ch = cin.get(); //get next character
        }
        if((Ch=='\n')&& (isalpha(cin.peek()))) //check for apostrophe in the word
        {
            ++LengthOfWord; //count apostrophe in word length
            Ch = cin.get(); //and get the next letter
        }
        if(isspace(Ch) || ispunct(Ch) || cin.eof()) //is it end of word
        {
            EndOfWord++;
        }
    }
    return LengthOfWord;
}


void DisplayFrequencyTable(const int Words[])
{
    int TotalWords = 0, TotalLength = 0;
    cout << "\nWord Length Frequency\n";
    cout << "------------ ----------\n";

    for (int i=1; i<=WORD_LENGTH-1; i++)
    {
        cout << setw(4)<<i<<setw(18)<<Words[i]<<endl;
        TotalLength += (i*Words[i]);
        TotalWords += Words[i];
    }
    cout << "\nAverage word length is ";

    if (TotalLength)
    {
        cout << float(TotalLength)/TotalWords << endl;
    }
    else cout << 0 << endl;
}

提前致谢。希望有人能帮忙。

Answer 1

尽管在 NextWordLength 中声明了 fin，但您的函数从不使用它。相反，它从 cin 读取，因此您的程序希望您输入文本以供其处理。