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

并不是说我们需要更多的答案,但这是我受到埃文·特兰启发后想到的。

std::vector <std::string> split(const string &input, auto delimiter, bool skipEmpty=true) {
  /*
  Splits a string at each delimiter and returns these strings as a string vector.
  If the delimiter is not found then nothing is returned.
  If skipEmpty is true then strings between delimiters that are 0 in length will be skipped.
  */
  bool delimiterFound = false;
  int pos=0, pPos=0;
  std::vector <std::string> result;
  while (true) {
    pos = input.find(delimiter,pPos);
    if (pos != std::string::npos) {
      if (skipEmpty==false or pos-pPos > 0) // if empty values are to be kept or not
        result.push_back(input.substr(pPos,pos-pPos));
      delimiterFound = true;
    } else {
      if (pPos < input.length() and delimiterFound) {
        if (skipEmpty==false or input.length()-pPos > 0) // if empty values are to be kept or not
          result.push_back(input.substr(pPos,input.length()-pPos));
      }
      break;
    }
    pPos = pos+1;
  }
  return result;
}

其他回答

使用std::string_view和Eric Niebler的range-v3库:

https://wandbox.org/permlink/kW5lwRCL1pxjp2pW

#include <iostream>
#include <string>
#include <string_view>
#include "range/v3/view.hpp"
#include "range/v3/algorithm.hpp"

int main() {
    std::string s = "Somewhere down the range v3 library";
    ranges::for_each(s  
        |   ranges::view::split(' ')
        |   ranges::view::transform([](auto &&sub) {
                return std::string_view(&*sub.begin(), ranges::distance(sub));
            }),
        [](auto s) {std::cout << "Substring: " << s << "\n";}
    );
}

通过使用循环的范围而不是范围::for_each算法:

#include <iostream>
#include <string>
#include <string_view>
#include "range/v3/view.hpp"

int main()
{
    std::string str = "Somewhere down the range v3 library";
    for (auto s : str | ranges::view::split(' ')
                      | ranges::view::transform([](auto&& sub) { return std::string_view(&*sub.begin(), ranges::distance(sub)); }
                      ))
    {
        std::cout << "Substring: " << s << "\n";
    }
}

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。

如果您希望按某些字符分割字符串,可以使用

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

using namespace std;
void replaceOtherChars(string &input, vector<char> &dividers)
{
    const char divider = dividers.at(0);
    int replaceIndex = 0;
    vector<char>::iterator it_begin = dividers.begin()+1,
        it_end= dividers.end();
    for(;it_begin!=it_end;++it_begin)
    {
        replaceIndex = 0;
        while(true)
        {
            replaceIndex=input.find_first_of(*it_begin,replaceIndex);
            if(replaceIndex==-1)
                break;
            input.at(replaceIndex)=divider;
        }
    }
}
vector<string> split(string str, vector<char> chars, bool missEmptySpace =true )
{
    vector<string> result;
    const char divider = chars.at(0);
    replaceOtherChars(str,chars);
    stringstream stream;
    stream<<str;    
    string temp;
    while(getline(stream,temp,divider))
    {
        if(missEmptySpace && temp.empty())
            continue;
        result.push_back(temp);
    }
    return result;
}
int main()
{
    string str ="milk, pigs.... hot-dogs ";
    vector<char> arr;
    arr.push_back(' '); arr.push_back(','); arr.push_back('.');
    vector<string> result = split(str,arr);
    vector<string>::iterator it_begin= result.begin(),
        it_end= result.end();
    for(;it_begin!=it_end;++it_begin)
    {
        cout<<*it_begin<<endl;
    }
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。

// adapted from a "regular" csv parse
std::string stringIn = "my csv  is 10233478 NOTseparated by commas";
std::vector<std::string> commaSeparated(1);
int commaCounter = 0;
for (int i=0; i<stringIn.size(); i++) {
    if (stringIn[i] == " ") {
        commaSeparated.push_back("");
        commaCounter++;
    } else {
        commaSeparated.at(commaCounter) += stringIn[i];
    }
}

最后你会得到一个字符串向量,句子中的每个元素都用空格隔开。唯一的非标准资源是std::vector(但由于涉及std::string,我认为它是可以接受的)。

空字符串保存为单独的项。