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

还有另一种方式——连续传递方式、零分配、基于函数的分隔。

 void split( auto&& data, auto&& splitter, auto&& operation ) {
   using std::begin; using std::end;
   auto prev = begin(data);
   while (prev != end(data) ) {
     auto&&[prev,next] = splitter( prev, end(data) );
     operation(prev,next);
     prev = next;
   }
 }

现在我们可以基于此编写特定的拆分函数。

 auto anyOfSplitter(auto delimiters) {
   return [delimiters](auto begin, auto end) {
     while( begin != end && 0 == std::string_view(begin, end).find_first_of(delimiters) ) {
       ++begin;
     }
     auto view = std::string_view(begin, end);
     auto next = view.find_first_of(delimiters);
     if (next != view.npos)
       return std::make_pair( begin, begin + next );
     else
       return std::make_pair( begin, end );
   };
 }

我们现在可以生成一个传统的std字符串分割,如下所示:

 template<class C>
 auto traditional_any_of_split( std::string_view<C> str, std::string_view<C> delim ) {
   std::vector<std::basic_string<C>> retval;
   split( str, anyOfSplitter(delim), [&](auto s, auto f) {
     retval.emplace_back(s,f);
   });
   return retval;
 }

或者我们可以改用find

 auto findSplitter(auto delimiter) {
   return [delimiter](auto begin, auto end) {
     while( begin != end && 0 == std::string_view(begin, end).find(delimiter) ) {
       begin += delimiter.size();
     }
     auto view = std::string_view(begin, end);
     auto next = view.find(delimiter);
     if (next != view.npos)
       return std::make_pair( begin, begin + next );
     else
       return std::make_pair( begin, end );
   };
 }

 template<class C>
 auto traditional_find_split( std::string_view<C> str, std::string_view<C> delim ) {
   std::vector<std::basic_string<C>> retval;
   split( str, findSplitter(delim), [&](auto s, auto f) {
     retval.emplace_back(s,f);
   });
   return retval;
 }

通过更换分流器部分。

这两者都分配了一个返回值缓冲区。我们可以以手动管理生命周期为代价将返回值交换到字符串视图。

我们还可以采用一个延续,一次传递一个字符串视图,甚至避免分配视图向量。

这可以通过一个中止选项进行扩展,这样我们可以在读取几个前缀字符串后中止。

其他回答

没有任何内存分配的C++17版本(std::函数除外)

void iter_words(const std::string_view& input, const std::function<void(std::string_view)>& process_word) {

    auto itr = input.begin();

    auto consume_whitespace = [&]() {
        for(; itr != input.end(); ++itr) {
            if(!isspace(*itr))
                return;
        }
    };

    auto consume_letters = [&]() {
        for(; itr != input.end(); ++itr) {
            if(isspace(*itr))
                return;
        }
    };

    while(true) {
        consume_whitespace();
        if(itr == input.end())
            return;
        auto word_start = itr - input.begin();
        consume_letters();
        auto word_end = itr - input.begin();
        process_word(input.substr(word_start, word_end - word_start));
    }
}

int main() {
    iter_words("foo bar", [](std::string_view sv) {
        std::cout << "Got word: " <<  sv << '\n';
    });
    return 0;
}

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;
}

我刚刚写了一个很好的例子,说明如何按符号拆分一个字符,然后将每个字符数组(由符号分隔的单词)放入一个向量中。为了简单起见,我创建了std字符串的向量类型。

我希望这对你有帮助,并且对你可读。

#include <vector>
#include <string>
#include <iostream>

void push(std::vector<std::string> &WORDS, std::string &TMP){
    WORDS.push_back(TMP);
    TMP = "";
}
std::vector<std::string> mySplit(char STRING[]){
        std::vector<std::string> words;
        std::string s;
        for(unsigned short i = 0; i < strlen(STRING); i++){
            if(STRING[i] != ' '){
                s += STRING[i];
            }else{
                push(words, s);
            }
        }
        push(words, s);//Used to get last split
        return words;
}

int main(){
    char string[] = "My awesome string.";
    std::cout << mySplit(string)[2];
    std::cin.get();
    return 0;
}

这类似于堆栈溢出问题:如何在C++中标记字符串?。需要Boost外部库

#include <iostream>
#include <string>
#include <boost/tokenizer.hpp>

using namespace std;
using namespace boost;

int main(int argc, char** argv)
{
    string text = "token  test\tstring";

    char_separator<char> sep(" \t");
    tokenizer<char_separator<char>> tokens(text, sep);
    for (const string& t : tokens)
    {
        cout << t << "." << endl;
    }
}

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));
}