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

// 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。

其他回答

我相信还没有人发布这个解决方案。与其直接使用分隔符,它基本上与boost::split()相同,即它允许您传递一个谓词,如果字符是分隔符,则返回true,否则返回false。我认为这给了程序员更多的控制,最棒的是你不需要提升。

template <class Container, class String, class Predicate>
void split(Container& output, const String& input,
           const Predicate& pred, bool trimEmpty = false) {
    auto it = begin(input);
    auto itLast = it;
    while (it = find_if(it, end(input), pred), it != end(input)) {
        if (not (trimEmpty and it == itLast)) {
            output.emplace_back(itLast, it);
        }
        ++it;
        itLast = it;
    }
}

然后可以这样使用:

struct Delim {
    bool operator()(char c) {
        return not isalpha(c);
    }
};    

int main() {
    string s("#include<iostream>\n"
             "int main() { std::cout << \"Hello world!\" << std::endl; }");

    vector<string> v;

    split(v, s, Delim(), true);
    /* Which is also the same as */
    split(v, s, [](char c) { return not isalpha(c); }, true);

    for (const auto& i : v) {
        cout << i << endl;
    }
}

这是我的版本

#include <vector>

inline std::vector<std::string> Split(const std::string &str, const std::string &delim = " ")
{
    std::vector<std::string> tokens;
    if (str.size() > 0)
    {
        if (delim.size() > 0)
        {
            std::string::size_type currPos = 0, prevPos = 0;
            while ((currPos = str.find(delim, prevPos)) != std::string::npos)
            {
                std::string item = str.substr(prevPos, currPos - prevPos);
                if (item.size() > 0)
                {
                    tokens.push_back(item);
                }
                prevPos = currPos + 1;
            }
            tokens.push_back(str.substr(prevPos));
        }
        else
        {
            tokens.push_back(str);
        }
    }
    return tokens;
}

它适用于多字符分隔符。它防止空令牌进入结果。它使用单个标头。当您不提供分隔符时,它将字符串作为一个标记返回。如果字符串为空,它还会返回一个空结果。不幸的是,它的效率很低,因为存在巨大的std::vector副本,除非您使用C++11进行编译,否则应该使用移动示意图。在C++11中,这段代码应该很快。

仅为方便:

template<class V, typename T>
bool in(const V &v, const T &el) {
    return std::find(v.begin(), v.end(), el) != v.end();
}

基于多个分隔符的实际拆分:

std::vector<std::string> split(const std::string &s,
                               const std::vector<char> &delims) {
    std::vector<std::string> res;
    auto stuff = [&delims](char c) { return !in(delims, c); };
    auto space = [&delims](char c) { return in(delims, c); };
    auto first = std::find_if(s.begin(), s.end(), stuff);
    while (first != s.end()) {
        auto last = std::find_if(first, s.end(), space);
        res.push_back(std::string(first, last));
        first = std::find_if(last + 1, s.end(), stuff);
    }
    return res;
}

用法:

int main() {
    std::string s = "   aaa,  bb  cc ";
    for (auto el: split(s, {' ', ','}))
        std::cout << el << std::endl;
    return 0;
}
void splitString(string str, char delim, string array[], const int arraySize)
{
    int delimPosition, subStrSize, subStrStart = 0;

    for (int index = 0; delimPosition != -1; index++)
    {
        delimPosition = str.find(delim, subStrStart);
        subStrSize = delimPosition - subStrStart;
        array[index] = str.substr(subStrStart, subStrSize);
        subStrStart =+ (delimPosition + 1);
    }
}

这个呢

#include <string>
#include <vector>

using namespace std;

vector<string> split(string str, const char delim) {
    vector<string> v;
    string tmp;

    for(string::const_iterator i; i = str.begin(); i <= str.end(); ++i) {
        if(*i != delim && i != str.end()) {
            tmp += *i; 
        } else {
            v.push_back(tmp);
            tmp = ""; 
        }   
    }   

    return v;
}