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

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";
    }
}

其他回答

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

int main() {
  string str = "ABC AABCD CDDD RABC GHTTYU FR";
  str += " "; //dirty hack: adding extra space to the end
  vector<string> v;

  for (int i=0; i<(int)str.size(); i++) {
    int a, b;
    a = i;

    for (int j=i; j<(int)str.size(); j++) {
      if (str[j] == ' ') {
        b = j;
        i = j;
        break;
      }
    }
    v.push_back(str.substr(a, b-a));
  }

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

值得一提的是,这里有另一种从输入字符串中提取令牌的方法,仅依赖于标准库设施。这是STL设计背后力量和优雅的一个例子。

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

int main() {
    using namespace std;
    string sentence = "And I feel fine...";
    istringstream iss(sentence);
    copy(istream_iterator<string>(iss),
         istream_iterator<string>(),
         ostream_iterator<string>(cout, "\n"));
}

可以使用相同的通用复制算法将提取的令牌插入到容器中,而不是将其复制到输出流中。

vector<string> tokens;
copy(istream_iterator<string>(iss),
     istream_iterator<string>(),
     back_inserter(tokens));

…或直接创建矢量:

vector<string> tokens{istream_iterator<string>{iss},
                      istream_iterator<string>{}};

有一个名为strtok的函数。

#include<string>
using namespace std;

vector<string> split(char* str,const char* delim)
{
    char* saveptr;
    char* token = strtok_r(str,delim,&saveptr);

    vector<string> result;

    while(token != NULL)
    {
        result.push_back(token);
        token = strtok_r(NULL,delim,&saveptr);
    }
    return result;
}

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", ':');
// 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。