如何计算C++字符数组中的单词

How to count words from an array of chars in C++

本文关键字：数组 字符 单词 C++ 何计算 计算      更新时间：2023-10-16

我在尝试将字符数组中的字母接收到我的wordCount函数中以计算数组中每个项目中的单词数时遇到问题。我相信我应该只操作该函数，但不清楚如何将 testCases 数组中的单个字母放入字数统计函数中。

之后，我假设我会使用 if 语句来检查读入 wordCount 的字符是否是字母，以及当它们结束时将它们计为一个单词。

代码如下：

#include <iostream>
using namespace std;
// Function Prototype
int wordCount (char *userEntry);
int main() {
// Constants
const int MAX_LENGTH = 150;
// Local variables
char testCases[][MAX_LENGTH + 1] = { "0",
    "    1   22    3333    44444 ",
    "     testing    ",
    "a",
    "onetwothree",
    "one two three",
    "    testing    a    11   222  three  4  five  ",
    "a b c d e f" };
int wCount = 0;
// loop through test cases and display number of words in each
for (char *entry : testCases) {
    wCount = wordCount(entry);
    cout << "nNumber of words in the test case '" << entry << "' is: " 
<< wCount << endl;
}
return EXIT_SUCCESS;
}
/*
Function Name:  wordCount
This function counts the # of space-delimited words
in a character string, and returns the count to the
caller.
NOTE: A word is defined as one or more alphabetic
characters separated by one or more spaces,
unless it is the only alphabetic character(s).
*/
int wordCount (char *userEntry) {
return 0;
}

c++ 中的字符串以 null 结尾，这意味着在它们的最后一个字符之后有一个字符代码0x00的''字符。要读取字符串/字符数组的每个字符，您只需使用索引运算符 [] 。与其他数组一样，C++中的字符串从 0 到 n-1 进行索引

下面是一个循环示例，它将字符串的字符读取到字符变量中。

void iterate_through_characters(const char* aString) {
    // starting at n=0 check each n and make sure it is shorter than the
    // width of the array
    // and that it's not the null terminating character
    for (int n = 0; (n < MAX_LENGTH + 1) && (aString[n] != ''); n++) {
        // take the character out of the index in the string and store it in aCharacter
        char aCharacter = aString[n];
    }
}

对于您的特殊情况，您还需要跟踪您是否已经在某个单词中，并且仅在您还没有在单词中时才计算一个新单词。下面的函数实现了这一点。

int wordCount(const char* input) {
    // this is true if we're in a word
    bool inWord = false; 
    // the number of words we've seen defaulting to 0, no words
    int result = 0;
    for (int n = 0; (n < MAX_LENGTH + 1) && (aString[n] != ''); n++) {
       // if this is a space we're not in a word
       if (aString[n] == ' ') {
           inWord = false; // if we were in a word, we aren't now
       } else if (!inWord) {
           inWord = true; // if we weren't in a word, we are now
           result ++; // increment the number of words we've seen 
       }
   }
   return result;
}
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