我知道你想要一个c++的解决方案，但你可能会认为这是有帮助的:

Qt

#include <QString>

...

QString str = "The quick brown fox"; 
QStringList results = str.split(" "); 

在这个例子中，与Boost相比的优势在于，它直接一对一地映射到你的文章代码。

详见Qt文档

您可以使用流、迭代器和复制算法来相当直接地做到这一点。

#include <string>
#include <vector>
#include <iostream>
#include <istream>
#include <ostream>
#include <iterator>
#include <sstream>
#include <algorithm>

int main()
{
  std::string str = "The quick brown fox";

  // construct a stream from the string
  std::stringstream strstr(str);

  // use stream iterators to copy the stream to the vector as whitespace separated strings
  std::istream_iterator<std::string> it(strstr);
  std::istream_iterator<std::string> end;
  std::vector<std::string> results(it, end);

  // send the vector to stdout.
  std::ostream_iterator<std::string> oit(std::cout);
  std::copy(results.begin(), results.end(), oit);
}

Boost有很强的拆分功能:Boost::algorithm::split。

示例程序:

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

int main() {
    auto s = "a,b, c ,,e,f,";
    std::vector<std::string> fields;
    boost::split(fields, s, boost::is_any_of(","));
    for (const auto& field : fields)
        std::cout << "\"" << field << "\"\n";
    return 0;
}

输出:

"a"
"b"
" c "
""
"e"
"f"
""

这里有许多过于复杂的建议。试试这个简单的std::string解决方案:

using namespace std;

string someText = ...

string::size_type tokenOff = 0, sepOff = tokenOff;
while (sepOff != string::npos)
{
    sepOff = someText.find(' ', sepOff);
    string::size_type tokenLen = (sepOff == string::npos) ? sepOff : sepOff++ - tokenOff;
    string token = someText.substr(tokenOff, tokenLen);
    if (!token.empty())
        /* do something with token */;
    tokenOff = sepOff;
}

使用regex_token_iterators的解决方案:

#include <iostream>
#include <regex>
#include <string>

using namespace std;

int main()
{
    string str("The quick brown fox");

    regex reg("\\s+");

    sregex_token_iterator iter(str.begin(), str.end(), reg, -1);
    sregex_token_iterator end;

    vector<string> vec(iter, end);

    for (auto a : vec)
    {
        cout << a << endl;
    }
}

我只是看了所有的答案，找不到下一个前提条件的解决方案:

没有动态内存分配 不使用boost 不使用正则表达式 c++17标准

这就是我的解

#include <iomanip>
#include <iostream>
#include <iterator>
#include <string_view>
#include <utility>

struct split_by_spaces
{
    std::string_view      text;
    static constexpr char delim = ' ';

    struct iterator
    {
        const std::string_view& text;
        std::size_t             cur_pos;
        std::size_t             end_pos;

        std::string_view operator*() const
        {
            return { &text[cur_pos], end_pos - cur_pos };
        }
        bool operator==(const iterator& other) const
        {
            return cur_pos == other.cur_pos && end_pos == other.end_pos;
        }
        bool operator!=(const iterator& other) const
        {
            return !(*this == other);
        }
        iterator& operator++()
        {
            cur_pos = text.find_first_not_of(delim, end_pos);

            if (cur_pos == std::string_view::npos)
            {
                cur_pos = text.size();
                end_pos = cur_pos;
                return *this;
            }

            end_pos = text.find(delim, cur_pos);

            if (cur_pos == std::string_view::npos)
            {
                end_pos = text.size();
            }

            return *this;
        }
    };

    [[nodiscard]] iterator begin() const
    {
        auto start = text.find_first_not_of(delim);
        if (start == std::string_view::npos)
        {
            return iterator{ text, text.size(), text.size() };
        }
        auto end_word = text.find(delim, start);
        if (end_word == std::string_view::npos)
        {
            end_word = text.size();
        }
        return iterator{ text, start, end_word };
    }
    [[nodiscard]] iterator end() const
    {
        return iterator{ text, text.size(), text.size() };
    }
};

int main(int argc, char** argv)
{
    using namespace std::literals;
    auto str = " there should be no memory allocation during parsing"
               "  into words this line and you   should'n create any"
               "  contaner                  for intermediate words  "sv;

    auto comma = "";
    for (std::string_view word : split_by_spaces{ str })
    {
        std::cout << std::exchange(comma, ",") << std::quoted(word);
    }

    auto only_spaces = "                   "sv;
    for (std::string_view word : split_by_spaces{ only_spaces })
    {
        std::cout << "you will not see this line in output" << std::endl;
    }
}