C++ 需要一些关于 Pig Latin 字符串的建议

Question

我需要用Pig Latin形式写一个句子，我几乎成功了，除了1个案例，我几乎放弃了 例如 ： 如果我的单词以 a\e\o\u\i 开头，这个词看起来就像 easy -> easyway , apple -> appleway

如果它不是以我上面写的字母开头 它看起来像这样：box -> oxbay , king -> ingkay

我成功使用了粗体部分，但在第一部分以 \e\o\u\i 开头的字母开头，我不知道将 w 放在哪里，需要一些帮助

这是我的代码，在此先感谢

#include <iostream>

//Since those are used in ALL function of program, it wont hurt to set it to global
//Else it is considered EVIL to declare global variables
const int maxLine = 100;
char phraseLine[maxLine] = { '\0' };

void pigLatinString();

using namespace std;


void main()
{
    // Displayed heading of program
    
    cout << "* You will be prompted to enter a string of *" << endl;
    cout << "* words. The string will be converted into  *" << endl;
    cout << "* Pig Latin and the results displayed.      *" << endl;
    cout << "* Enter as many strings as you would like.  *" << endl;
    

    //prompt the user for a group of words or press enter to quit
    cout << "Please enter a word or group of words. (Press enter to quit)\n";
    cin.getline(phraseLine, 100, '\n');
    cout << endl;

    // This is the main loop. Continue executing until the user hits 'enter' to quit.
    while (phraseLine[0] != '\0')
    {

        // Display the word (s) entered by the user
        cout << "You entered the following: " << phraseLine << endl;

        // Display the word (s) in Pig Latin
        cout << "The same phrase in Pig latin is: ";
        pigLatinString();
        cout << endl;

        //prompt the user for a group of words or press enter to quit
        cout << "Please enter a word or group of words. (Press enter to quit)\n";
        cin.getline(phraseLine, 100, '\n');
    }
    return;
}


void pigLatinString() //phraseLine is a cstring for the word,  maxline is max length of line
{ //variable declarations
    char tempConsonant[10];
    tempConsonant[0] = '\0';
    int numberOfConsonants = 0;
    char previousCharacter = ' ';
    char currentCharacter = ' ';
    bool isInWord = 0;

    // for loop checking each index to the end of whatever is typed in
    for (int i = 0; i < maxLine; i++)
    {

        //checking for the end of the phraseline 
        if (phraseLine[i] == '\0')
        {//checking to see if it's in the word
            if (isInWord)
            {//checking to see that there wasn't a space ahead of the word and then sending the cstring + ay to the console
                if (previousCharacter != ' ')
                    cout << tempConsonant << "ay" << endl;
            }
            return;
        }

        // this covers the end of the word condition
        if (isInWord)
        {// covers the condition of index [i] being the space at the end of the word
            if (phraseLine[i] == ' ')
            {
                // spits out pig latin word, gets you out of the word, flushes the temp consonants array and resets the # of consonants to 0
                cout << tempConsonant << "ay";
                isInWord = 0;
                tempConsonant[0] = '\0';
                numberOfConsonants = 0;
            }
            cout << phraseLine[i] ;

        }
        else
        {//this covers for the first vowel that makes the switch
            if (phraseLine[i] != ' ')
            {// sets the c string to what is in the phraseline at the time and makes it capitalized
                char currentCharacter = phraseLine[i];
                currentCharacter = toupper(currentCharacter);

                // this takes care of the condition that currentCharacter is not a vowel
                if ((currentCharacter != 'A') && (currentCharacter != 'E') &&
                    (currentCharacter != 'I') && (currentCharacter != 'O') && (currentCharacter != 'U'))
                    //this sets the array to temporarily hold the consonants for display before the 'ay'
                {//this sets the null operator at the end of the c string and looks for the next consonant
                    tempConsonant[numberOfConsonants] = phraseLine[i];
                    tempConsonant[numberOfConsonants + 1] = '\0';
                    numberOfConsonants++;
                }
                else
                {// this sets the boolean isInWord to true and displays the phraseline
                    isInWord = 1;
                    cout << phraseLine[i];
                }
            }
            else
            {
                cout << phraseLine[i] ;
            }
        }

        previousCharacter = phraseLine[i];
    }
    return;
}

Answer 1

您需要考虑两个条件。 if您的单词以元音开头，只需在单词末尾添加“way”，else移动第一个字母并在末尾添加“ay”。

使用std::string 代替C 字符串可以使这项任务变得更简单。这是因为您现在不再担心超出长度或丢失空字符。它还允许更轻松地访问标准库算法。

#include <algorithm>
#include <iostream>
#include <string>

std::string make_pig_latin(const std::string& word) {
  std::string vowels("aeiou");
  std::string newWord(word);

  if (newWord.find_first_not_of(vowels) == 0) {
    // Word starts with a consanant
    std::rotate(newWord.begin(), newWord.begin() + 1, newWord.end());
    newWord += "ay";
  } else {
    newWord += "way";
  }

  return newWord;
}

int main() {
  std::cout << make_pig_latin("apple") << '\n'
            << make_pig_latin("box") << '\n'
            << make_pig_latin("king") << '\n'
            << make_pig_latin("easy") << '\n';
}

上面的函数重点介绍了如何构建转化结构。您只需要知道您的单词是否以元音开头，然后采取适当的措施。

输出：

appleway
oxbay
ingkay
easyway

我没有得到你必须关心“电话”之类的词的印象。

查看您的代码，您应该尝试更好地分离您的关注点。 Pig Latin 更容易一次完成一个单词，但是您的 Pig Latin 函数中有字符串拆分代码和许多“非 Pig Latin”代码。您的 main 可以处理获取输入。您可能应该有一个单独的函数来将行分解为单个单词，最好使用std::vector 来保存单词，因为它可以按需增长，并且不必预先知道特定的容量。然后，您遍历您的单词数组并单独翻译它们。根据您的实际要求，您甚至可能不必存储翻译后的单词，只需将它们直接打印到屏幕上即可。

这是相同的程序，但现在它可以分隔单词。注意 pig latin 函数不需要更改（很多，我添加了大写元音只是因为我不想费心转换单词）以便添加添加的功能。

#include <algorithm>
#include <iostream>
#include <iterator>
#include <sstream>
#include <string>
#include <vector>

std::string make_pig_latin(const std::string& word) {
  std::string vowels("aeiouAEIOU");
  std::string newWord(word);

  if (newWord.find_first_not_of(vowels) == 0) {
    // Word starts with a consanant
    std::rotate(newWord.begin(), newWord.begin() + 1, newWord.end());
    newWord += "ay";
  } else {
    newWord += "way";
  }

  return newWord;
}

int main() {
  std::string phrase(
      "A sentence where I say words like apple box easy king and ignore "
      "punctuation");

  std::istringstream sin(phrase);
  std::vector<std::string> words(std::istream_iterator<std::string>(sin), {});

  for (auto i : words) {
    std::cout << make_pig_latin(i) << ' ';
  }
}

输出：

Away entencesay hereway Iway aysay ordsway ikelay appleway oxbay easyway ingkay andway ignoreway unctuationpay