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

下面是一个更好的方法。它可以接受任何字符,除非您愿意,否则不会拆分行。不需要特殊的库(嗯,除了std,但谁真的认为这是一个额外的库),没有指针,没有引用,而且它是静态的。只是简单的C++。

#pragma once
#include <vector>
#include <sstream>
using namespace std;
class Helpers
{
    public:
        static vector<string> split(string s, char delim)
        {
            stringstream temp (stringstream::in | stringstream::out);
            vector<string> elems(0);
            if (s.size() == 0 || delim == 0)
                return elems;
            for(char c : s)
            {
                if(c == delim)
                {
                    elems.push_back(temp.str());
                    temp = stringstream(stringstream::in | stringstream::out);
                }
                else
                    temp << c;
            }
            if (temp.str().size() > 0)
                elems.push_back(temp.str());
                return elems;
            }

        //Splits string s with a list of delimiters in delims (it's just a list, like if we wanted to
        //split at the following letters, a, b, c we would make delims="abc".
        static vector<string> split(string s, string delims)
        {
            stringstream temp (stringstream::in | stringstream::out);
            vector<string> elems(0);
            bool found;
            if(s.size() == 0 || delims.size() == 0)
                return elems;
            for(char c : s)
            {
                found = false;
                for(char d : delims)
                {
                    if (c == d)
                    {
                        elems.push_back(temp.str());
                        temp = stringstream(stringstream::in | stringstream::out);
                        found = true;
                        break;
                    }
                }
                if(!found)
                    temp << c;
            }
            if(temp.str().size() > 0)
                elems.push_back(temp.str());
            return elems;
        }
};

其他回答

我的代码是:

#include <list>
#include <string>
template<class StringType = std::string, class ContainerType = std::list<StringType> >
class DSplitString:public ContainerType
{
public:
    explicit DSplitString(const StringType& strString, char cChar, bool bSkipEmptyParts = true)
    {
        size_t iPos = 0;
        size_t iPos_char = 0;
        while(StringType::npos != (iPos_char = strString.find(cChar, iPos)))
        {
            StringType strTemp = strString.substr(iPos, iPos_char - iPos);
            if((bSkipEmptyParts && !strTemp.empty()) || (!bSkipEmptyParts))
                push_back(strTemp);
            iPos = iPos_char + 1;
        }
    }
    explicit DSplitString(const StringType& strString, const StringType& strSub, bool bSkipEmptyParts = true)
    {
        size_t iPos = 0;
        size_t iPos_char = 0;
        while(StringType::npos != (iPos_char = strString.find(strSub, iPos)))
        {
            StringType strTemp = strString.substr(iPos, iPos_char - iPos);
            if((bSkipEmptyParts && !strTemp.empty()) || (!bSkipEmptyParts))
                push_back(strTemp);
            iPos = iPos_char + strSub.length();
        }
    }
};

例子:

#include <iostream>
#include <string>
int _tmain(int argc, _TCHAR* argv[])
{
    DSplitString<> aa("doicanhden1;doicanhden2;doicanhden3;", ';');
    for each (std::string var in aa)
    {
        std::cout << var << std::endl;
    }
    std::cin.get();
    return 0;
}

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;
}

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());

我一直在寻找用任意长度的分隔符分割字符串的方法,所以我开始从头开始编写,因为现有的解决方案不适合我。

这是我的小算法,仅使用STL:

//use like this
//std::vector<std::wstring> vec = Split<std::wstring> (L"Hello##world##!", L"##");

template <typename valueType>
static std::vector <valueType> Split (valueType text, const valueType& delimiter)
{
    std::vector <valueType> tokens;
    size_t pos = 0;
    valueType token;

    while ((pos = text.find(delimiter)) != valueType::npos) 
    {
        token = text.substr(0, pos);
        tokens.push_back (token);
        text.erase(0, pos + delimiter.length());
    }
    tokens.push_back (text);

    return tokens;
}

根据我的测试,它可以与任何长度和形式的分隔符一起使用。使用string或wstring类型实例化。

该算法所做的就是搜索分隔符,获取字符串中与分隔符相邻的部分,删除分隔符,然后再次搜索,直到不再找到为止。

当然,可以使用任意数量的空格作为分隔符。

我希望这有帮助。

有一种更简单的方法可以做到这一点!!

#include <vector>
#include <string>
std::vector<std::string> splitby(std::string string, char splitter) {
    int splits = 0;
    std::vector<std::string> result = {};
    std::string locresult = "";
    for (unsigned int i = 0; i < string.size(); i++) {
        if ((char)string.at(i) != splitter) {
            locresult += string.at(i);
        }
        else {
            result.push_back(locresult);
            locresult = "";
        }
    }
    if (splits == 0) {
        result.push_back(locresult);
    }
    return result;
}

void printvector(std::vector<std::string> v) {
    std::cout << '{';
    for (unsigned int i = 0; i < v.size(); i++) {
        if (i < v.size() - 1) {
            std::cout << '"' << v.at(i) << "\",";
        }
        else {
            std::cout << '"' << v.at(i) << "\"";
        }
    }
    std::cout << "}\n";
}