C-使用find(）分解字符串

我正在编写一个pig拉丁文转换器，它将字符串作为命令行参数，并将句子中的每个单词转换为pig拉丁文。我试图通过查找空格将字符串分解为单独的单词，但我用下面的代码得到的输出。。。老实说毫无意义。

它适用于第一个单词，但不会转换任何后续单词。所以"测试这个字符串"变成了"esttay this ay string ay"

在这一点上，我们还没有涵盖向量，所以解决方案必须相对简单。

有什么想法吗？

#include <iostream>
#include <string>
using namespace std;
void convertSentence(string sentence);
string pigLatinWord(string input);
bool isVowel(string input);
int main(int argc, char* argv[]) {
    if(argc != 2) {
        cout << "USAGE: " << argv[0] << " "[sentence]"" << endl;
    } else {
        string sentence(argv[1]);
        convertSentence(sentence);
    }
    return 0;
}
void convertSentence(string sentence) {
    string word = "", remainder = sentence, pigLatin = "";
    for(int i = sentence.find(" ",0); i != string::npos; i = sentence.find(" ",i)) {
        word = sentence.substr(0,i);
        pigLatin += pigLatinWord(word) + " ";
        sentence = sentence.erase(0,i);
        i++;
    }
    pigLatin += pigLatinWord(sentence);
    cout << pigLatin << endl;
}

string pigLatinWord(string input) {
    string word = "";
// If the first letter is a vowel, simply add "yay" to the end of it.
       if (isVowel(input)) {
            word = input + "yay";
//If the first letter is a consonant, add the first letter to the end,
//delete the first letter, then add "ay." - Credit to Tyler Sorrels
//CString/String solution post on Piazza. I had issues working with
//substrings, and this ended up being easier to read.
//But I did add a check for words that start with two consonants
//to cover all cases.
    } else {
        input += input.at(0);
        input = input.erase(0,1);
        word = input + "ay";
    }
    return word;
}
// Checks if the first letter of the word is a vowel.
// Returns true if yes, false if no.
bool isVowel(string input) {
     return ((input[0] == 'a') || (input[0] == 'e') || (input[0] == 'i') || (input[0]      == 'o') || (input[0] == 'a'));
}