如何按字符拆分字符串？C++

Question

我正在做一个有趣的国际民航组织翻译程序，我几乎完成了，但我有一个小问题。如何按每个字符分割字符串？例如，我想要的输出是；在国际民航组织字母表中翻译的单词mike是：

m:迈克

i:印第安纳州

k:公斤

e:回声

到目前为止，我才知道，国际民航组织字母中的mike一词是：

麦克

印第安纳州

基洛

回声

显然，我的文章主要是代码，我必须增加更多细节，所以我添加这个句子希望满足需求。此外，翻译应该是在对方之上，而不是一个额外的空间下来。我对此有问题，我想知道如何解决这个问题。

#include <iostream>
#include <string>

using namespace std;

int main()
{
   string word = " ", phonetic;
   int count = 0;
   
   cout << "Enter a word: ";
   cin >> word;
   
   while(count < word.length())
    {
        switch(word.at(count))
        {
            case 'A': case 'a': phonetic += " Alpha\n";
                break;
            case 'B': case 'b': phonetic += " Bravo\n";
                break;
            case 'C': case 'c': phonetic += " Charlie\n";
                break;
            case 'D': case 'd': phonetic += " Delta\n";
                break;
            case 'E': case 'e': phonetic += " Echo\n";
                break;
            case 'F': case 'f': phonetic += " Foxtrot\n";
                break;
            case 'G': case 'g': phonetic += " Golf\n";
                break;
            case 'H': case 'h': phonetic += " Hotel\n";
                break;
            case 'I': case 'i': phonetic += " Indiana\n";
                break;
            case 'J': case 'j': phonetic += " Juliet\n";
                break;
            case 'K': case 'k': phonetic += " Kilo\n";
                break;
            case 'L': case 'l': phonetic += " Lima\n";
                break;
            case 'M': case 'm': phonetic += " Mike\n";
                break;
            case 'N': case 'n': phonetic += " November\n";
                break;
            case 'O': case 'o': phonetic += " Oscar\n";
                break;
            case 'P': case 'p': phonetic += " Papa\n";
                break;
            case 'Q': case 'q': phonetic += " Quebec\n";
                break;
            case 'R': case 'r': phonetic += " Romeo\n";
                break;
            case 'S': case 's': phonetic += " Sierra\n";
                break;
            case 'T': case 't': phonetic += " Tango\n";
                break;
            case 'U': case 'u': phonetic += " Uniform\n";
                break;
            case 'V': case 'v': phonetic += " Victor\n";
                break;
            case 'W': case 'w': phonetic += " Whiskey\n";
                break;
            case 'X': case 'x': phonetic += " X-Ray\n";
                break;
            case 'Y': case 'y': phonetic += " Yankee\n";
                break;
            case 'Z': case 'z': phonetic += " Zulu\n";
                break;
            default: cout << "You did not enter a name" << endl;
            
        }
        count++;
    }

    cout << "The word "<< word <<" in the ICAO alphabet is:\n" 
    << phonetic << endl;
    
    return 0;
}

Answer 1

如果我正确理解您的帖子，您不希望拆分字符串，而是迭代字符串中的字符。

在C++11中：

for (char& c : word) {
    // switch (c)
}

Answer 2

要遍历字符串，只需使用迭代器：

std::string test_string = "test";
for( auto const& character : test_string )
{
    std::cout << character << "\n";
}

通过使用地图，可以简化整个程序：

// Example program
#include <iostream>
#include <map>
#include <string>

char to_lower(char ch)
{
    return static_cast<char>(std::tolower(static_cast<unsigned char>(ch)));
}

int main()
{
    const std::map<char, std::string> phonetic_alphabet = 
    {
         {'a', "Alpha"}
        ,{'b', "Bravo"}
        ,{'c', "Charlie"}
        ,{'d', "Delta"}
        ,{'e', "Echo"}
        ,{'f', "Foxtrot"}
        ,{'g', "Golf"}
        ,{'h', "Hotel"}
        ,{'i', "Indiana"}
        ,{'j', "Julia"}
        ,{'k', "Kilo"}
        ,{'l', "Lima"}
        ,{'m', "Mike"}
        ,{'n', "November"}
        ,{'o', "Oscar"}
        ,{'p', "Papa"}
        ,{'q', "Quebec"}
        ,{'r', "Romeo"}
        ,{'s', "Sierra"}
        ,{'t', "Tango"}
        ,{'u', "Uniform"}
        ,{'v', "Victor"}
        ,{'w', "Whiskey"}
        ,{'x', "X-Ray"}
        ,{'y', "Yankee"}
        ,{'z', "Zulu"}
    };
    
  std::cout << "Enter a word: ";
  std::string word;
  std::cin >> word;
  for( auto const& c : word )
  {
      const char lower_c = to_lower(c);
        if( phonetic_alphabet.find(lower_c) != phonetic_alphabet.end() )
        {
            std::cout << phonetic_alphabet.at(lower_c) << " ";
        }
        else
        {
            std::cout << c << " ";
        }
  }
}

Answer 3

无穷小零的答案是完全可以的，在这种情况下可能是最好的解决方案。

我想附加显示一个无循环的解决方案，也是基于一个std::unordered_map，并能够阅读完整的句子。

基本思想是:基于将单个字符转换为ICAOwords的要求，我决定为此使用一个专用函数：std::transform。您可能会看到文档这里

请参阅下文的补充/备选解决方案：

#include <iostream>
#include <unordered_map>
#include <string>
#include <algorithm>
#include <iterator>

int main()
{
    // Preinitialize an unordered map
    std::unordered_map<int, std::string> alpha =
    {
        {'a', "Alpha"},{'b', "Bravo"},{'c', "Charlie"},{'d', "Delta"},{'e', "Echo"},{'f', "Foxtrot"},{'g', "Golf"},
        {'h', "Hotel"},{'i', "Indiana"},{'j', "Julia"},{'k', "Kilo"},{'l', "Lima"} ,{'m', "Mike"},{'n', "November"},
        {'o', "Oscar"},{'p', "Papa"},{'q', "Quebec"},{'r', "Romeo"},{'s', "Sierra"},{'t', "Tango"},{'u', "Uniform"},
        {'v', "Victor"},{'w', "Whiskey"},{'x', "X-Ray"},{'y', "Yankee"},{'z', "Zulu"}};

    // Read word form user
    if (std::string word; std::getline(std::cin, word)) {

        // Show output to user
        std::cout << "\n\n" << word << " --> ";

        // Convert and show
        std::transform(word.begin(), word.end(), std::ostream_iterator<std::string>(std::cout, " "),
            [&](const char c) { return (isalpha(c) ? alpha[tolower(c)] : ""); });
    }
    return 0;
}