如何迭代由空格分隔的单词组成的字符串中的单词?

注意,我对C字符串函数或那种字符操作/访问不感兴趣。比起效率,我更喜欢优雅。我当前的解决方案:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int main() {
    string s = "Somewhere down the road";
    istringstream iss(s);

    do {
        string subs;
        iss >> subs;
        cout << "Substring: " << subs << endl;
    } while (iss);
}

当前回答

如果您喜欢使用boost,但希望使用整个字符串作为分隔符(而不是之前提出的大多数解决方案中的单个字符),可以使用boost_split_iterator。

示例代码包括方便的模板:

#include <iostream>
#include <vector>
#include <boost/algorithm/string.hpp>

template<typename _OutputIterator>
inline void split(
    const std::string& str, 
    const std::string& delim, 
    _OutputIterator result)
{
    using namespace boost::algorithm;
    typedef split_iterator<std::string::const_iterator> It;

    for(It iter=make_split_iterator(str, first_finder(delim, is_equal()));
            iter!=It();
            ++iter)
    {
        *(result++) = boost::copy_range<std::string>(*iter);
    }
}

int main(int argc, char* argv[])
{
    using namespace std;

    vector<string> splitted;
    split("HelloFOOworldFOO!", "FOO", back_inserter(splitted));

    // or directly to console, for example
    split("HelloFOOworldFOO!", "FOO", ostream_iterator<string>(cout, "\n"));
    return 0;
}

其他回答

我有一种与其他解决方案非常不同的方法,它提供了很多其他解决方案所缺乏的价值,但当然也有其缺点。这是一个工作实现,示例是在单词周围放置<tag></tag>。

首先,这个问题可以通过一个循环解决,不需要额外的内存,只需考虑四种逻辑情况。从概念上讲,我们对边界感兴趣。我们的代码应该反映出这一点:让我们遍历字符串,一次查看两个字符,记住字符串的开头和结尾都有特殊情况。

缺点是我们必须编写实现,这有点冗长,但大多是方便的样板。

好处是我们编写了实现,因此很容易根据特定的需要定制它,例如区分左和写单词边界,使用任何一组分隔符,或处理其他情况,例如无边界或错误位置。

using namespace std;

#include <iostream>
#include <string>

#include <cctype>

typedef enum boundary_type_e {
    E_BOUNDARY_TYPE_ERROR = -1,
    E_BOUNDARY_TYPE_NONE,
    E_BOUNDARY_TYPE_LEFT,
    E_BOUNDARY_TYPE_RIGHT,
} boundary_type_t;

typedef struct boundary_s {
    boundary_type_t type;
    int pos;
} boundary_t;

bool is_delim_char(int c) {
    return isspace(c); // also compare against any other chars you want to use as delimiters
}

bool is_word_char(int c) {
    return ' ' <= c && c <= '~' && !is_delim_char(c);
}

boundary_t maybe_word_boundary(string str, int pos) {
    int len = str.length();
    if (pos < 0 || pos >= len) {
        return (boundary_t){.type = E_BOUNDARY_TYPE_ERROR};
    } else {
        if (pos == 0 && is_word_char(str[pos])) {
            // if the first character is word-y, we have a left boundary at the beginning
            return (boundary_t){.type = E_BOUNDARY_TYPE_LEFT, .pos = pos};
        } else if (pos == len - 1 && is_word_char(str[pos])) {
            // if the last character is word-y, we have a right boundary left of the null terminator
            return (boundary_t){.type = E_BOUNDARY_TYPE_RIGHT, .pos = pos + 1};
        } else if (!is_word_char(str[pos]) && is_word_char(str[pos + 1])) {
            // if we have a delimiter followed by a word char, we have a left boundary left of the word char
            return (boundary_t){.type = E_BOUNDARY_TYPE_LEFT, .pos = pos + 1};
        } else if (is_word_char(str[pos]) && !is_word_char(str[pos + 1])) {
            // if we have a word char followed by a delimiter, we have a right boundary right of the word char
            return (boundary_t){.type = E_BOUNDARY_TYPE_RIGHT, .pos = pos + 1};
        }
        return (boundary_t){.type = E_BOUNDARY_TYPE_NONE};
    }
}

int main() {
    string str;
    getline(cin, str);

    int len = str.length();
    for (int i = 0; i < len; i++) {
        boundary_t boundary = maybe_word_boundary(str, i);
        if (boundary.type == E_BOUNDARY_TYPE_LEFT) {
            // whatever
        } else if (boundary.type == E_BOUNDARY_TYPE_RIGHT) {
            // whatever
        }
    }
}

正如您所看到的,代码非常容易理解和微调,代码的实际使用非常简短和简单。使用C++不应阻止我们编写最简单、最容易定制的代码,即使这意味着不使用STL。我认为这是Linus Torvalds所说的“品味”的一个例子,因为我们已经消除了所有不需要的逻辑,而写作风格自然允许在需要处理的时候处理更多的案件。

可以改进此代码的可能是使用enum类,在maybe_word_boundary中接受指向is_word_char的函数指针,而不是直接调用is_word_char,并传递lambda。

最小的解决方案是一个函数,它将std::字符串和一组分隔符(作为std::string)作为输入,并返回std:::字符串的std::向量。

#include <string>
#include <vector>

std::vector<std::string>
tokenize(const std::string& str, const std::string& delimiters)
{
  using ssize_t = std::string::size_type;
  const ssize_t str_ln = str.length();
  ssize_t last_pos = 0;

  // container for the extracted tokens
  std::vector<std::string> tokens;

  while (last_pos < str_ln) {
      // find the position of the next delimiter
      ssize_t pos = str.find_first_of(delimiters, last_pos);

      // if no delimiters found, set the position to the length of string
      if (pos == std::string::npos)
         pos = str_ln;

      // if the substring is nonempty, store it in the container
      if (pos != last_pos)
         tokens.emplace_back(str.substr(last_pos, pos - last_pos));

      // scan past the previous substring
      last_pos = pos + 1;
  }

  return tokens;
}

用法示例:

#include <iostream>

int main()
{
    std::string input_str = "one + two * (three - four)!!---! ";
    const char* delimiters = "! +- (*)";
    std::vector<std::string> tokens = tokenize(input_str, delimiters);

    std::cout << "input = '" << input_str << "'\n"
              << "delimiters = '" << delimiters << "'\n"
              << "nr of tokens found = " << tokens.size() << std::endl;
    for (const std::string& tk : tokens) {
        std::cout << "token = '" << tk << "'\n";
    }

  return 0;
}

仅为方便:

template<class V, typename T>
bool in(const V &v, const T &el) {
    return std::find(v.begin(), v.end(), el) != v.end();
}

基于多个分隔符的实际拆分:

std::vector<std::string> split(const std::string &s,
                               const std::vector<char> &delims) {
    std::vector<std::string> res;
    auto stuff = [&delims](char c) { return !in(delims, c); };
    auto space = [&delims](char c) { return in(delims, c); };
    auto first = std::find_if(s.begin(), s.end(), stuff);
    while (first != s.end()) {
        auto last = std::find_if(first, s.end(), space);
        res.push_back(std::string(first, last));
        first = std::find_if(last + 1, s.end(), stuff);
    }
    return res;
}

用法:

int main() {
    std::string s = "   aaa,  bb  cc ";
    for (auto el: split(s, {' ', ','}))
        std::cout << el << std::endl;
    return 0;
}

对于那个些需要使用字符串分隔符拆分字符串的人,也许可以尝试我的以下解决方案。

std::vector<size_t> str_pos(const std::string &search, const std::string &target)
{
    std::vector<size_t> founds;

    if(!search.empty())
    {
        size_t start_pos = 0;

        while (true)
        {
            size_t found_pos = target.find(search, start_pos);

            if(found_pos != std::string::npos)
            {
                size_t found = found_pos;

                founds.push_back(found);

                start_pos = (found_pos + 1);
            }
            else
            {
                break;
            }
        }
    }

    return founds;
}

std::string str_sub_index(size_t begin_index, size_t end_index, const std::string &target)
{
    std::string sub;

    size_t size = target.length();

    const char* copy = target.c_str();

    for(size_t i = begin_index; i <= end_index; i++)
    {
        if(i >= size)
        {
            break;
        }
        else
        {
            char c = copy[i];

            sub += c;
        }
    }

    return sub;
}

std::vector<std::string> str_split(const std::string &delimiter, const std::string &target)
{
    std::vector<std::string> splits;

    if(!delimiter.empty())
    {
        std::vector<size_t> founds = str_pos(delimiter, target);

        size_t founds_size = founds.size();

        if(founds_size > 0)
        {
            size_t search_len = delimiter.length();

            size_t begin_index = 0;

            for(int i = 0; i <= founds_size; i++)
            {
                std::string sub;

                if(i != founds_size)
                {
                    size_t pos  = founds.at(i);

                    sub = str_sub_index(begin_index, pos - 1, target);

                    begin_index = (pos + search_len);
                }
                else
                {
                    sub = str_sub_index(begin_index, (target.length() - 1), target);
                }

                splits.push_back(sub);
            }
        }
    }

    return splits;
}

这些片段由3个函数组成。坏消息是使用str_split函数,您将需要另外两个函数。是的,这是一大块代码。但好消息是,这两个附加功能可以独立工作,有时也很有用

测试main()块中的函数如下:

int main()
{
    std::string s = "Hello, world! We need to make the world a better place. Because your world is also my world, and our children's world.";

    std::vector<std::string> split = str_split("world", s);

    for(int i = 0; i < split.size(); i++)
    {
        std::cout << split[i] << std::endl;
    }
}

它将产生:

Hello, 
! We need to make the 
 a better place. Because your 
 is also my 
, and our children's 
.

我认为这不是最有效的代码,但至少它可以工作。希望有帮助。

我们可以在c++中使用strtok,

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
    char str[]="Mickey M;12034;911416313;M;01a;9001;NULL;0;13;12;0;CPP,C;MSC,3D;FEND,BEND,SEC;";
    char *pch = strtok (str,";,");
    while (pch != NULL)
    {
        cout<<pch<<"\n";
        pch = strtok (NULL, ";,");
    }
    return 0;
}