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

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;
}

其他回答

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。

我用这个分隔符分隔字符串。第一个将结果放入预先构建的向量中,第二个返回新向量。

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

template <typename Out>
void split(const std::string &s, char delim, Out result) {
    std::istringstream iss(s);
    std::string item;
    while (std::getline(iss, item, delim)) {
        *result++ = item;
    }
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, std::back_inserter(elems));
    return elems;
}

请注意,此解决方案不会跳过空令牌,因此下面将找到4项,其中一项为空:

std::vector<std::string> x = split("one:two::three", ':');

这类似于堆栈溢出问题:如何在C++中标记字符串?。需要Boost外部库

#include <iostream>
#include <string>
#include <boost/tokenizer.hpp>

using namespace std;
using namespace boost;

int main(int argc, char** argv)
{
    string text = "token  test\tstring";

    char_separator<char> sep(" \t");
    tokenizer<char_separator<char>> tokens(text, sep);
    for (const string& t : tokens)
    {
        cout << t << "." << endl;
    }
}

有一种更简单的方法可以做到这一点!!

#include <vector>
#include <string>
std::vector<std::string> splitby(std::string string, char splitter) {
    int splits = 0;
    std::vector<std::string> result = {};
    std::string locresult = "";
    for (unsigned int i = 0; i < string.size(); i++) {
        if ((char)string.at(i) != splitter) {
            locresult += string.at(i);
        }
        else {
            result.push_back(locresult);
            locresult = "";
        }
    }
    if (splits == 0) {
        result.push_back(locresult);
    }
    return result;
}

void printvector(std::vector<std::string> v) {
    std::cout << '{';
    for (unsigned int i = 0; i < v.size(); i++) {
        if (i < v.size() - 1) {
            std::cout << '"' << v.at(i) << "\",";
        }
        else {
            std::cout << '"' << v.at(i) << "\"";
        }
    }
    std::cout << "}\n";
}

下面的代码使用strtok()将字符串拆分为标记,并将标记存储在向量中。

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

using namespace std;


char one_line_string[] = "hello hi how are you nice weather we are having ok then bye";
char seps[]   = " ,\t\n";
char *token;



int main()
{
   vector<string> vec_String_Lines;
   token = strtok( one_line_string, seps );

   cout << "Extracting and storing data in a vector..\n\n\n";

   while( token != NULL )
   {
      vec_String_Lines.push_back(token);
      token = strtok( NULL, seps );
   }
     cout << "Displaying end result in vector line storage..\n\n";

    for ( int i = 0; i < vec_String_Lines.size(); ++i)
    cout << vec_String_Lines[i] << "\n";
    cout << "\n\n\n";


return 0;
}