template<typename C, typename T>
auto insert_in_container(C& c, T&& t) -> decltype(c.push_back(std::forward<T>(t)), void()) {
    c.push_back(std::forward<T>(t));
}
template<typename C, typename T>
auto insert_in_container(C& c, T&& t) -> decltype(c.insert(std::forward<T>(t)), void()) {
    c.insert(std::forward<T>(t));
}
template<typename Container>
Container splitR(const std::string& input, const std::string& delims) {
    Container out;
    size_t delims_len = delims.size();
    auto begIdx = 0u;
    auto endIdx = input.find(delims, begIdx);
    if (endIdx == std::string::npos && input.size() != 0u) {
        insert_in_container(out, input);
    }
    else {
        size_t w = 0;
        while (endIdx != std::string::npos) {
            w = endIdx - begIdx;
            if (w != 0) insert_in_container(out, input.substr(begIdx, w));
            begIdx = endIdx + delims_len;
            endIdx = input.find(delims, begIdx);
        }
        w = input.length() - begIdx;
        if (w != 0) insert_in_container(out, input.substr(begIdx, w));
    }
    return out;
}

这是一个简洁的分裂函数。我决定让背靠背分隔符返回为空字符串，但您可以很容易地检查子字符串是否为空，如果是，则不将其添加到向量。

#include <vector>
#include <string>
using namespace std;



vector<string> split(string to_split, string delimiter) {
    size_t pos = 0;
    vector<string> matches{};
    do {
        pos = to_split.find(delimiter);
        int change_end;
        if (pos == string::npos) {
            pos = to_split.length() - 1;
            change_end = 1;
        }
        else {
            change_end = 0;
        }
        matches.push_back(to_split.substr(0, pos+change_end));
        
        to_split.erase(0, pos+1);

    }
    while (!to_split.empty());
    return matches;

}

这是一个完整的方法，它在任何分隔符上分割字符串，并返回分割后的字符串的向量。

这是改编自ryanbwork的答案。然而，他的检查:if(token != mystring)给出错误的结果，如果你的字符串中有重复的元素。这是我对那个问题的解决方案。

vector<string> Split(string mystring, string delimiter)
{
    vector<string> subStringList;
    string token;
    while (true)
    {
        size_t findfirst = mystring.find_first_of(delimiter);
        if (findfirst == string::npos) //find_first_of returns npos if it couldn't find the delimiter anymore
        {
            subStringList.push_back(mystring); //push back the final piece of mystring
            return subStringList;
        }
        token = mystring.substr(0, mystring.find_first_of(delimiter));
        mystring = mystring.substr(mystring.find_first_of(delimiter) + 1);
        subStringList.push_back(token);
    }
    return subStringList;
}

如果你不想修改字符串(就像Vincenzo Pii回答的那样)，并且想输出最后一个令牌，你可能想使用这种方法:

inline std::vector<std::string> splitString( const std::string &s, const std::string &delimiter ){
    std::vector<std::string> ret;
    size_t start = 0;
    size_t end = 0;
    size_t len = 0;
    std::string token;
    do{ end = s.find(delimiter,start); 
        len = end - start;
        token = s.substr(start, len);
        ret.emplace_back( token );
        start += len + delimiter.length();
        std::cout << token << std::endl;
    }while ( end != std::string::npos );
    return ret;
}

这段代码从文本中分离行，并将每个行添加到一个向量中。

vector<string> split(char *phrase, string delimiter){
    vector<string> list;
    string s = string(phrase);
    size_t pos = 0;
    string token;
    while ((pos = s.find(delimiter)) != string::npos) {
        token = s.substr(0, pos);
        list.push_back(token);
        s.erase(0, pos + delimiter.length());
    }
    list.push_back(s);
    return list;
}

调用:

vector<string> listFilesMax = split(buffer, "\n");

std::vector<std::string> split(const std::string& s, char c) {
  std::vector<std::string> v;
  unsigned int ii = 0;
  unsigned int j = s.find(c);
  while (j < s.length()) {
    v.push_back(s.substr(i, j - i));
    i = ++j;
    j = s.find(c, j);
    if (j >= s.length()) {
      v.push_back(s.substr(i, s,length()));
      break;
    }
  }
  return v;
}