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

没有Boost,没有字符串流,只有标准的C库与std::string和std::list:C库函数配合使用,便于分析,C++数据类型便于内存管理。

空白被认为是换行符、制表符和空格的任意组合。空白字符集由wschars变量建立。

#include <string>
#include <list>
#include <iostream>
#include <cstring>

using namespace std;

const char *wschars = "\t\n ";

list<string> split(const string &str)
{
  const char *cstr = str.c_str();
  list<string> out;

  while (*cstr) {                     // while remaining string not empty
    size_t toklen;
    cstr += strspn(cstr, wschars);    // skip leading whitespace
    toklen = strcspn(cstr, wschars);  // figure out token length
    if (toklen)                       // if we have a token, add to list
      out.push_back(string(cstr, toklen));
    cstr += toklen;                   // skip over token
  }

  // ran out of string; return list

  return out;
}

int main(int argc, char **argv)
{
  list<string> li = split(argv[1]);
  for (list<string>::iterator i = li.begin(); i != li.end(); i++)
    cout << "{" << *i << "}" << endl;
  return 0;
}

Run:

$ ./split ""
$ ./split "a"
{a}
$ ./split " a "
{a}
$ ./split " a b"
{a}
{b}
$ ./split " a b c"
{a}
{b}
{c}
$ ./split " a b c d  "
{a}
{b}
{c}
{d}

split的尾部递归版本(本身分裂为两个函数)。除了将字符串推入列表之外,所有对变量的破坏性操作都消失了!

void split_rec(const char *cstr, list<string> &li)
{
  if (*cstr) {
    const size_t leadsp = strspn(cstr, wschars);
    const size_t toklen = strcspn(cstr + leadsp, wschars);

    if (toklen)
      li.push_back(string(cstr + leadsp, toklen));

    split_rec(cstr + leadsp + toklen, li);
  }
}

list<string> split(const string &str)
{
  list<string> out;
  split_rec(str.c_str(), out);
  return out;
}

其他回答

对于一个大得离谱而且可能是冗余的版本,可以尝试很多For循环。

string stringlist[10];
int count = 0;

for (int i = 0; i < sequence.length(); i++)
{
    if (sequence[i] == ' ')
    {
        stringlist[count] = sequence.substr(0, i);
        sequence.erase(0, i+1);
        i = 0;
        count++;
    }
    else if (i == sequence.length()-1)  // Last word
    {
        stringlist[count] = sequence.substr(0, i+1);
    }
}

它并不漂亮,但总的来说(除了标点符号和一系列其他错误)它是有效的!

使用std::stringstream非常好,并且完全符合您的要求。如果您只是在寻找不同的方法,那么可以使用std::find()/std::find_first_of()和std::string::substr()。

下面是一个示例:

#include <iostream>
#include <string>

int main()
{
    std::string s("Somewhere down the road");
    std::string::size_type prev_pos = 0, pos = 0;

    while( (pos = s.find(' ', pos)) != std::string::npos )
    {
        std::string substring( s.substr(prev_pos, pos-prev_pos) );

        std::cout << substring << '\n';

        prev_pos = ++pos;
    }

    std::string substring( s.substr(prev_pos, pos-prev_pos) ); // Last word
    std::cout << substring << '\n';

    return 0;
}

我对string和u32string~的一般实现,使用boost::algorithm::split签名。

template<typename CharT, typename UnaryPredicate>
void split(std::vector<std::basic_string<CharT>>& split_result,
           const std::basic_string<CharT>& s,
           UnaryPredicate predicate)
{
    using ST = std::basic_string<CharT>;
    using std::swap;
    std::vector<ST> tmp_result;
    auto iter = s.cbegin(),
         end_iter = s.cend();
    while (true)
    {
        /**
         * edge case: empty str -> push an empty str and exit.
         */
        auto find_iter = find_if(iter, end_iter, predicate);
        tmp_result.emplace_back(iter, find_iter);
        if (find_iter == end_iter) { break; }
        iter = ++find_iter; 
    }
    swap(tmp_result, split_result);
}


template<typename CharT>
void split(std::vector<std::basic_string<CharT>>& split_result,
           const std::basic_string<CharT>& s,
           const std::basic_string<CharT>& char_candidate)
{
    std::unordered_set<CharT> candidate_set(char_candidate.cbegin(),
                                            char_candidate.cend());
    auto predicate = [&candidate_set](const CharT& c) {
        return candidate_set.count(c) > 0U;
    };
    return split(split_result, s, predicate);
}

template<typename CharT>
void split(std::vector<std::basic_string<CharT>>& split_result,
           const std::basic_string<CharT>& s,
           const CharT* literals)
{
    return split(split_result, s, std::basic_string<CharT>(literals));
}

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>{}};

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;
}