在在C文件计数的话 [英] Counting words in a file in C

问题描述

我正在写计数在一个文件中的字的数量的函数。词可以通过空格字符的数量分开。可存在于一个文件中的整数，但该方案应当只计算具有至少一个字母字符的话。

I'm writing a function that counts the number of words in a file. Words may be separated by any amount of whitespace characters. There can be integers in a file, but the program should only count words which have at least one alphabetic character.

int word_count(const char *filename)
{
    int ch;
    int state;
    int count = 0;
    FILE *fileHandle;
    if ((fileHandle = fopen(filename, "r")) == NULL){
        return -1;
    }

    state = OUT;
    count = 0;
    while ((ch = fgetc(fileHandle)) != EOF){
        if (isspace(ch))
            state = OUT;
        else if (state == OUT){
            state = IN;
            ++count;
        }
    }

    fclose(fileHandle);

    return count;  

}

我想通了，如何处理空格，但我不知道该怎么不能算不至少有一个字母字符（我知道因而isalpha和ISDIGIT，但我有困难，了解如何使用组合他们在我的情况）。

I figured out how to deal with whitespaces, but I don't know how not to count combinations which don't have at least one alphabetic character (I know about isalpha and isdigit, but I have difficulty in understanding how to use them in my case).

我真的AP preciate你的帮助。

I would really appreciate your help.

推荐答案

您只需更换：

else if (state == OUT){

与

else if (state == OUT && isalpha(ch)){

所以，你在第一个字符集的状态，并将其视作字。
要知道，你算 last.First 作为一个单词，可以考虑使用（！字符isalnum（CH））而不是（isspace为（CH））。

So you set the state to IN at the first character and count it as word. Be aware that you count last.First as a single word, consider using (!isalnum(ch)) instead of (isspace(ch)).

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