如何计算C++中的元音？

我的一个作业有问题。我的任务如下：

用C++编写程序 输入：字符串 输出：相邻元音的数量 示例：输入->计算机；输出->12因为c、mp、t、rs 我已经试过几次了，但都不管用。 这是我的一个尝试：

cout << "Type your word: ";
cin >> YourWord;
cout << "Your word is: " << YourWord << endl;

//maganhangzók
char a = 'a';
char e = 'e';
char i = 'i';
char o = 'o';
char u = 'u';

//massalhangzok
char b = 'b';
char c = 'c';
char d = 'd';
char f = 'f';
char g = 'g';
char h = 'h';
char j = 'j';
char k = 'k';
char l = 'l';
char m = 'm';
char n = 'n';
char p = 'p';
char q = 'q';
char r = 'r';
char s = 's';
char t = 't';
char v = 'v';
char w = 'w';
char x = 'x';
char y = 'y';
char z = 'z';

int counter[YourWord.length()];
int nothing = 0;

for(int i=1; i<YourWord.length(); i++) {
    if (((YourWord[i] = a) || (YourWord[i] = e) || (YourWord[i] =  i) || (YourWord[i] = o) || (YourWord[i] = u)) && ((YourWord[i-1] = a) || (YourWord[i-1] = e) || (YourWord[i-1] =  i) || (YourWord[i-1] = o) || (YourWord[i-1] = u))) {
        nothing++;
    } else if (((YourWord[i] = a) || (YourWord[i] = e) || (YourWord[i] =  i) || (YourWord[i] = o) || (YourWord[i] = u)) && ((YourWord[i-1] = b) || (YourWord[i-1] = c) || (YourWord[i-1] = d) || (YourWord[i-1] = f) || (YourWord[i-1] = g) || (YourWord[i-1] = j) || (YourWord[i-1] = k) || (YourWord[i-1] = l) || (YourWord[i-1] = m) || (YourWord[i-1] = n) || (YourWord[i-1] = p) || (YourWord[i-1] =  q) || (YourWord[i-1] = r) || (YourWord[i-1] = s) || (YourWord[i-1] = t) || (YourWord[i-1] =  v) || (YourWord[i-1] =  w) || (YourWord[i-1] = x) || (YourWord[i-1] = y) || (YourWord[i-1] = z))) {
        counter[i] = counter[i];
    } else if (((YourWord[i] = b) || (YourWord[i] = c) || (YourWord[i] =  d) || (YourWord[i] = f) || (YourWord[i] = g) || (YourWord[i] = j) || (YourWord[i] =  k) || (YourWord[i] = l) || (YourWord[i] = m) || (YourWord[i] = n) || (YourWord[i] = p) || (YourWord[i] = q) || (YourWord[i] = r) || (YourWord[i] = s) || (YourWord[i] = t) || (YourWord[i] = v) || (YourWord[i] =  w) || (YourWord[i] = x) || (YourWord[i] = y) || (YourWord[i] = z)) && ((YourWord[i-1] = b) || (YourWord[i-1] = c) || (YourWord[i-1] =  d) || (YourWord[i-1] = f) || (YourWord[i-1] = g) || (YourWord[i-1] = j) || (YourWord[i-1] = k) || (YourWord[i-1] = l) || (YourWord[i-1] = m) || (YourWord[i-1] = n) || (YourWord[i-1] = p) || (YourWord[i-1] = q) || (YourWord[i-1] = r) || (YourWord[i-1] = s) || (YourWord[i-1] = t) || (YourWord[i-1] =  v) || (YourWord[i-1] =  w) || (YourWord[i-1] = x) || (YourWord[i-1] = y) || (YourWord[i-1] = z))) {
        counter[i]++;
    } else {
            counter[i+1]++;
    }
}

不幸的是，这是不必要的复杂和绝对无法使用。

---

谁能帮我一个提示，我应该从哪里开始？

您的代码有一个问题，它似乎试图同时做两件事：

确定字符是否位于所选集合中 根据这些决定，计算请求的数字。 为了可读性，第2部分应该使用第1部分，而不是包括它

当您使用很长的代码行时，情况更是如此。因此，读者必须使用水平滑块，在这种情况下，不可能同时看到长代码行的结尾和随后的短代码行。这使得查找bug变得更加困难

<>这是C++，因此面向对象的编程，我们可以委托一个字符是否是所选的一部分到一个特定的对象，例如一个字符测试类的决定。ChartTester对象在创建时接收包含所有选定字符的字符串。这样，计算请求的数字的算法就可以短得多

ChartTester对象可以将所有繁重的工作委托给字符串

创建此辅助对象的源代码如下所示：

    static const char  EnglishConsonantList[] =
        "bcdfghjklmnpqrstvwxzBCDFGHJKLMNPQRSTVWXZ";
    CharTester  cht(EnglishConsonantList);

上述编码风格的一个好处是，如果某个官方语法委员会决定从现在起，Y将是辅音而不是元音，那么源代码中所需的更改是最小的。你的老师可能不会高高在上，比如«既然你已经为辅音做了，请为元音做。»-这是为了迫使您认识到源代码的灵活性

因此，我们将看到下面的代码，其中最有趣的部分是goodCharCounts函数，它返回一个包含请求的数字的std:：vector对象

#include  <vector>
#include  <string>
#include  <iostream>

using  std::string;
using  std::cout;


class CharTester {
public:
    CharTester(const string& list) : goodCharList(list)
    {};
    bool isGoodChar(char ch) const;

private:
    string  goodCharList;    // list of "chosen" characters
};

bool CharTester::isGoodChar(char ch) const
{
    // hard work there:
    bool isGood = (goodCharList.find(ch) != string::npos);

    return isGood;
}


// THE ALGORITHM:

std::vector<int>  goodCharCounts(const CharTester& cht, const std::string& str)
{
    std::vector<int>  posVec;

    int  counter = 0;
    // loop on all characters of the input string
    for (char ch : str) {
        bool isGood = cht.isGoodChar(ch);

        if (isGood) {
            counter++;
        }
        else if (counter > 0) {
            // end of current "chosen" group, so must register it
            posVec.push_back(counter);
            counter = 0;
        }
    }

    if (counter > 0) {
        // register last "chosen" group
        posVec.push_back(counter);
        counter = 0;
    }

    return  posVec;
}

旁注：

---

以上，我使用了char类型和相关的经典C++字符串类型，因为这是你在代码中的方式。但是，对于生产代码，您可能必须使用更通用的宽wchar__t和wstring类型，以便允许非ASCII字符，例如á、á、ê，ẞ., 等Unicode字符。

y呢？那不是元音吗？猜它不是匈牙利语的元音。方便阅读：YourWord[i]=a将a的值赋给YourWord[i]。YourWord[i]==a比较这两个值。在你的问题中，你说的是输出彼此相邻的元音数。然而，似乎你指的是辅音的数量，而不是元音。你可以使用包含所有元音或所有辅音的字符串变量listOfGoodChars，正如上面Andreas所暗示的。所有元音的列表都是数据而不是代码！然后使用findstring方法的表单4，如下所述：；like pos=listOfGoodChars.findch；如果位置！=是的，这是一个很好的性格。。。

// test one string and print the results:
void doUnitTest(const CharTester& cht, const string& str)
{
    std::vector<int>  counts = goodCharCounts(cht, str);

    cout << str << " --> ";
    // loop on all numbers:
    for (int n : counts) {
        cout << n << ' ';
    }
    cout << std::endl;
}


// possible lists of "chosen" characters :

//-- const wchar_t FrenchVowelList[]      = L"AEIOUYÀÂÉÊÈÏÔÛÙaeiouyàâéêèïôûù";
//-- const wchar_t RussianVowelList[]     = L"АЕИОЭЙЫЮЯЁУаеиоэйыюяёу";

static const char  EnglishVowelList[] =  "aeiouyAEIOUY";
static const char  EnglishConsonantList[] =
    "bcdfghjklmnpqrstvwxzBCDFGHJKLMNPQRSTVWXZ";


int main()
{
    CharTester  cht(EnglishConsonantList);  

    doUnitTest(cht, "computers");
    doUnitTest(cht, "Some computers are cheap. Aeiuo-XrZT.");

    return EXIT_SUCCESS;
}

computers --> 1 2 1 2 
Some computers are cheap. Aeiuo-XrZT. --> 1 1 1 2 1 2 1 2 1 4