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

#include <iostream>
#include <string>
#include <deque>

std::deque<std::string> split(
    const std::string& line, 
    std::string::value_type delimiter,
    bool skipEmpty = false
) {
    std::deque<std::string> parts{};

    if (!skipEmpty && !line.empty() && delimiter == line.at(0)) {
        parts.push_back({});
    }

    for (const std::string::value_type& c : line) {
        if (
            (
                c == delimiter 
                &&
                (skipEmpty ? (!parts.empty() && !parts.back().empty()) : true)
            )
            ||
            (c != delimiter && parts.empty())
        ) {
            parts.push_back({});
        }

        if (c != delimiter) {
            parts.back().push_back(c);
        }
    }

    if (skipEmpty && !parts.empty() && parts.back().empty()) {
        parts.pop_back();
    }

    return parts;
}

void test(const std::string& line) {
    std::cout << line << std::endl;

    std::cout << "skipEmpty=0 |";
    for (const std::string& part : split(line, ':')) {
        std::cout << part << '|';
    }
    std::cout << std::endl;

    std::cout << "skipEmpty=1 |";
    for (const std::string& part : split(line, ':', true)) {
        std::cout << part << '|';
    }
    std::cout << std::endl;

    std::cout << std::endl;
}

int main() {
    test("foo:bar:::baz");
    test("");
    test("foo");
    test(":");
    test("::");
    test(":foo");
    test("::foo");
    test(":foo:");
    test(":foo::");

    return 0;
}

输出:

foo:bar:::baz
skipEmpty=0 |foo|bar|||baz|
skipEmpty=1 |foo|bar|baz|


skipEmpty=0 |
skipEmpty=1 |

foo
skipEmpty=0 |foo|
skipEmpty=1 |foo|

:
skipEmpty=0 |||
skipEmpty=1 |

::
skipEmpty=0 ||||
skipEmpty=1 |

:foo
skipEmpty=0 ||foo|
skipEmpty=1 |foo|

::foo
skipEmpty=0 |||foo|
skipEmpty=1 |foo|

:foo:
skipEmpty=0 ||foo||
skipEmpty=1 |foo|

:foo::
skipEmpty=0 ||foo|||
skipEmpty=1 |foo|

其他回答

这是我使用C++11和STL的解决方案。它应该是合理有效的:

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

std::vector<std::string> split(const std::string& s)
{
    std::vector<std::string> v;

    const auto end = s.end();
    auto to = s.begin();
    decltype(to) from;

    while((from = std::find_if(to, end,
        [](char c){ return !std::isspace(c); })) != end)
    {
        to = std::find_if(from, end, [](char c){ return std::isspace(c); });
        v.emplace_back(from, to);
    }

    return v;
}

int main()
{
    std::string s = "this is the string  to  split";

    auto v = split(s);

    for(auto&& s: v)
        std::cout << s << '\n';
}

输出:

this
is
the
string
to
split

LazyString拆分器:

#include <string>
#include <algorithm>
#include <unordered_set>

using namespace std;

class LazyStringSplitter
{
    string::const_iterator start, finish;
    unordered_set<char> chop;

public:

    // Empty Constructor
    explicit LazyStringSplitter()
    {}

    explicit LazyStringSplitter (const string cstr, const string delims)
        : start(cstr.begin())
        , finish(cstr.end())
        , chop(delims.begin(), delims.end())
    {}

    void operator () (const string cstr, const string delims)
    {
        chop.insert(delims.begin(), delims.end());
        start = cstr.begin();
        finish = cstr.end();
    }

    bool empty() const { return (start >= finish); }

    string next()
    {
        // return empty string
        // if ran out of characters
        if (empty())
            return string("");

        auto runner = find_if(start, finish, [&](char c) {
            return chop.count(c) == 1;
        });

        // construct next string
        string ret(start, runner);
        start = runner + 1;

        // Never return empty string
        // + tail recursion makes this method efficient
        return !ret.empty() ? ret : next();
    }
};

我将此方法称为LazyStringSplitter是因为一个原因——它不会一次性拆分字符串。本质上,它的行为类似于python生成器它公开了一个名为next的方法,该方法返回从原始字符串拆分的下一个字符串我使用了c++11STL中的无序集,因此查找分隔符的速度要快得多下面是它的工作原理

测试程序

#include <iostream>
using namespace std;

int main()
{
    LazyStringSplitter splitter;

    // split at the characters ' ', '!', '.', ','
    splitter("This, is a string. And here is another string! Let's test and see how well this does.", " !.,");

    while (!splitter.empty())
        cout << splitter.next() << endl;
    return 0;
}

输出,输出

This
is
a
string
And
here
is
another
string
Let's
test
and
see
how
well
this
does

改进这一点的下一个计划是实施开始和结束方法,以便可以执行以下操作:

vector<string> split_string(splitter.begin(), splitter.end());
#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;
}

最小的解决方案是一个函数,它将std::字符串和一组分隔符(作为std::string)作为输入,并返回std:::字符串的std::向量。

#include <string>
#include <vector>

std::vector<std::string>
tokenize(const std::string& str, const std::string& delimiters)
{
  using ssize_t = std::string::size_type;
  const ssize_t str_ln = str.length();
  ssize_t last_pos = 0;

  // container for the extracted tokens
  std::vector<std::string> tokens;

  while (last_pos < str_ln) {
      // find the position of the next delimiter
      ssize_t pos = str.find_first_of(delimiters, last_pos);

      // if no delimiters found, set the position to the length of string
      if (pos == std::string::npos)
         pos = str_ln;

      // if the substring is nonempty, store it in the container
      if (pos != last_pos)
         tokens.emplace_back(str.substr(last_pos, pos - last_pos));

      // scan past the previous substring
      last_pos = pos + 1;
  }

  return tokens;
}

用法示例:

#include <iostream>

int main()
{
    std::string input_str = "one + two * (three - four)!!---! ";
    const char* delimiters = "! +- (*)";
    std::vector<std::string> tokens = tokenize(input_str, delimiters);

    std::cout << "input = '" << input_str << "'\n"
              << "delimiters = '" << delimiters << "'\n"
              << "nr of tokens found = " << tokens.size() << std::endl;
    for (const std::string& tk : tokens) {
        std::cout << "token = '" << tk << "'\n";
    }

  return 0;
}

短而优雅

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

vector<string> split(string data, string token)
{
    vector<string> output;
    size_t pos = string::npos; // size_t to avoid improbable overflow
    do
    {
        pos = data.find(token);
        output.push_back(data.substr(0, pos));
        if (string::npos != pos)
            data = data.substr(pos + token.size());
    } while (string::npos != pos);
    return output;
}

可以使用任何字符串作为分隔符,也可以与二进制数据一起使用(std::string支持二进制数据,包括空值)

使用:

auto a = split("this!!is!!!example!string", "!!");

输出:

this
is
!example!string