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

虽然有一些答案提供了C++20解决方案,但自从发布以来,已经做了一些更改,并将其作为缺陷报告应用于C++20。正因为如此,解决方案变得更短、更好:

#include <iostream>
#include <ranges>
#include <string_view>

namespace views = std::views;
using str = std::string_view;

constexpr str text = "Lorem ipsum dolor sit amet, consectetur adipiscing elit.";

auto splitByWords(str input) {
    return input
    | views::split(' ')
    | views::transform([](auto &&r) -> str {
        return {r.begin(), r.end()};
    });
}

auto main() -> int {
    for (str &&word : splitByWords(text)) {
        std::cout << word << '\n';
    }
}

到今天为止,它仍然只在GCC的主干分支(Godbolt链接)上可用。它基于两个更改:P1391迭代器构造函数用于std::string_view和P2210 DR修复std::views::split以保留范围类型。

在C++23中,不需要任何转换样板,因为P1989向std::string_view:添加了一个范围构造函数

#include <iostream>
#include <ranges>
#include <string_view>

namespace views = std::views;

constexpr std::string_view text = "Lorem ipsum dolor sit amet, consectetur adipiscing elit.";

auto main() -> int {
    for (std::string_view&& word : text | views::split(' ')) {
        std::cout << word << '\n';
    }
}

(螺栓连杆)

获取Boost!:-)

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

using namespace std;
using namespace boost;

int main(int argc, char**argv) {
    typedef vector < string > list_type;

    list_type list;
    string line;

    line = "Somewhere down the road";
    split(list, line, is_any_of(" "));

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

    return 0;
}

此示例给出了输出-

Somewhere
down
the
road

到目前为止,我在Boost中使用了这个,但我需要一些不依赖它的东西,所以我得出了这个结论:

static void Split(std::vector<std::string>& lst, const std::string& input, const std::string& separators, bool remove_empty = true)
{
    std::ostringstream word;
    for (size_t n = 0; n < input.size(); ++n)
    {
        if (std::string::npos == separators.find(input[n]))
            word << input[n];
        else
        {
            if (!word.str().empty() || !remove_empty)
                lst.push_back(word.str());
            word.str("");
        }
    }
    if (!word.str().empty() || !remove_empty)
        lst.push_back(word.str());
}

好的一点是,在分隔符中可以传递多个字符。

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

    vector<string> split(const string &s, char delim) {
        vector<string> elems;
        stringstream ss(s);
        string item;
        while (getline(ss, item, delim)) {
            elems.push_back(item);
        }
        return elems;
    }

int main() {

        vector<string> x = split("thi is an sample test",' ');
        unsigned int i;
        for(i=0;i<x.size();i++)
            cout<<i<<":"<<x[i]<<endl;
        return 0;
}