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

#include <vector>
#include <string>
#include <sstream>

int main()
{
    std::string str("Split me by whitespaces");
    std::string buf;                 // Have a buffer string
    std::stringstream ss(str);       // Insert the string into a stream

    std::vector<std::string> tokens; // Create vector to hold our words

    while (ss >> buf)
        tokens.push_back(buf);

    return 0;
}

其他回答

这是我写的一个函数,帮助我做了很多事情。它在为WebSocket做协议时帮助了我。

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

vector<string> split ( string input , string split_id ) {
  vector<string> result;
  int i = 0;
  bool add;
  string temp;
  stringstream ss;
  size_t found;
  string real;
  int r = 0;
    while ( i != input.length() ) {
        add = false;
        ss << input.at(i);
        temp = ss.str();
        found = temp.find(split_id);
        if ( found != string::npos ) {
            add = true;
            real.append ( temp , 0 , found );
        } else if ( r > 0 &&  ( i+1 ) == input.length() ) {
            add = true;
            real.append ( temp , 0 , found );
        }
        if ( add ) {
            result.push_back(real);
            ss.str(string());
            ss.clear();
            temp.clear();
            real.clear();
            r = 0;
        }
        i++;
        r++;
    }
  return result;
}

int main() {
    string s = "S,o,m,e,w,h,e,r,e, down the road \n In a really big C++ house.  \n  Lives a little old lady.   \n   That no one ever knew.    \n    She comes outside.     \n     In the very hot sun.      \n\n\n\n\n\n\n\n   And throws C++ at us.    \n    The End.  FIN.";
    vector < string > Token;
    Token = split ( s , "," );
    for ( int i = 0 ; i < Token.size(); i++)    cout << Token.at(i) << endl;
    cout << endl << Token.size();
    int a;
    cin >> a;
    return a;
}

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;
}

使用Boost的可能解决方案可能是:

#include <boost/algorithm/string.hpp>
std::vector<std::string> strs;
boost::split(strs, "string to split", boost::is_any_of("\t "));

这种方法可能比字符串流方法更快。由于这是一个通用模板函数,因此可以使用各种分隔符拆分其他类型的字符串(wchar等或UTF-8)。

有关详细信息,请参阅文档。

最近我不得不将一个骆驼大小写的单词拆分成子单词。没有分隔符,只有大写字符。

#include <string>
#include <list>
#include <locale> // std::isupper

template<class String>
const std::list<String> split_camel_case_string(const String &s)
{
    std::list<String> R;
    String w;

    for (String::const_iterator i = s.begin(); i < s.end(); ++i) {  {
        if (std::isupper(*i)) {
            if (w.length()) {
                R.push_back(w);
                w.clear();
            }
        }
        w += *i;
    }

    if (w.length())
        R.push_back(w);
    return R;
}

例如,这将“AQueryTrades”拆分为“A”、“Query”和“Trades”。该函数适用于窄字符串和宽字符串。因为它尊重当前的语言环境,所以将“RaumfahrtÜberwachungsVerordnung”分为“Raumfahrt”、“Überwachungs”和“Verordnug”。

注意std::upper应该真正作为函数模板参数传递。然后,此函数的更广义的from也可以在分隔符(如“、”、“;”或“”)处拆分。

我有两条线来解决这个问题:

char sep = ' ';
std::string s="1 This is an example";

for(size_t p=0, q=0; p!=s.npos; p=q)
  std::cout << s.substr(p+(p!=0), (q=s.find(sep, p+1))-p-(p!=0)) << std::endl;

然后你可以把它放到一个向量中,而不是打印。