c - 如何计算字符串 C 程序中的单词和标点符号？

Question

我正在尝试用 C 语言编写一个程序，该程序计算字符串中单词和标点符号的数量，而不使用数组等内置函数。没有数组可以这样做吗？另外，我当前的程序在下面，并给我一个初始化 *word 的错误，但我试图让用户输入一个字符串并且程序对其进行计数，所以我不想初始化。非常感谢您的帮助!

    #include <stdio.h>
    #include<conio.h>

    int main(){
        char *word;
        int countword = 0, i;
        int countpunct = 0, i;
        printf("\nEnter the String: ");
        gets(word);
        for (i = 0; word[i] == ' '; i++){
            countword++;
        }
        for (i = 0; word[i] == '.' || '?' || '!' || '(' || ')' || '*' || '&'){
            countpunct++;
        }
        printf("\nThe number of words is %d.", countword);
        printf("\nThe number of punctuation marsks is %d.", countpunct);
        getch();

    }

Answer 1

一种方法是分别读取每个字符并进行处理。

#include <stdio.h>
#if 0
#include<conio.h>
#endif

int main(){
    int word;
    int countword = 0;
    int countpunct = 0;
    printf("\nEnter the String: ");
    while ((word = getchar()) != EOF && word != '\n'){
        if (word == ' ') countword++;
        if (word == '.' || word == '?' ||  word == '!' ||  word == '(' ||  word == ')' ||  word == '*' ||  word == '&'){
            countpunct++;
        }
    }
    printf("\nThe number of words is %d.", countword);
    printf("\nThe number of punctuation marsks is %d.", countpunct);
#if 0
    getch();
#endif
}