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

如果您喜欢使用boost,但希望使用整个字符串作为分隔符(而不是之前提出的大多数解决方案中的单个字符),可以使用boost_split_iterator。

示例代码包括方便的模板:

#include <iostream>
#include <vector>
#include <boost/algorithm/string.hpp>

template<typename _OutputIterator>
inline void split(
    const std::string& str, 
    const std::string& delim, 
    _OutputIterator result)
{
    using namespace boost::algorithm;
    typedef split_iterator<std::string::const_iterator> It;

    for(It iter=make_split_iterator(str, first_finder(delim, is_equal()));
            iter!=It();
            ++iter)
    {
        *(result++) = boost::copy_range<std::string>(*iter);
    }
}

int main(int argc, char* argv[])
{
    using namespace std;

    vector<string> splitted;
    split("HelloFOOworldFOO!", "FOO", back_inserter(splitted));

    // or directly to console, for example
    split("HelloFOOworldFOO!", "FOO", ostream_iterator<string>(cout, "\n"));
    return 0;
}

其他回答

到目前为止,我在Boost中使用了这个,但我需要一些不依赖它的东西,所以我得出了这个结论:

static void Split(std::vector<std::string>& lst, const std::string& input, const std::string& separators, bool remove_empty = true)
{
    std::ostringstream word;
    for (size_t n = 0; n < input.size(); ++n)
    {
        if (std::string::npos == separators.find(input[n]))
            word << input[n];
        else
        {
            if (!word.str().empty() || !remove_empty)
                lst.push_back(word.str());
            word.str("");
        }
    }
    if (!word.str().empty() || !remove_empty)
        lst.push_back(word.str());
}

好的一点是,在分隔符中可以传递多个字符。

获取Boost!:-)

#include <boost/algorithm/string/split.hpp>
#include <boost/algorithm/string.hpp>
#include <iostream>
#include <vector>

using namespace std;
using namespace boost;

int main(int argc, char**argv) {
    typedef vector < string > list_type;

    list_type list;
    string line;

    line = "Somewhere down the road";
    split(list, line, is_any_of(" "));

    for(int i = 0; i < list.size(); i++)
    {
        cout << list[i] << endl;
    }

    return 0;
}

此示例给出了输出-

Somewhere
down
the
road

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>{}};

我的实施可以是另一种解决方案:

std::vector<std::wstring> SplitString(const std::wstring & String, const std::wstring & Seperator)
{
    std::vector<std::wstring> Lines;
    size_t stSearchPos = 0;
    size_t stFoundPos;
    while (stSearchPos < String.size() - 1)
    {
        stFoundPos = String.find(Seperator, stSearchPos);
        stFoundPos = (stFoundPos == std::string::npos) ? String.size() : stFoundPos;
        Lines.push_back(String.substr(stSearchPos, stFoundPos - stSearchPos));
        stSearchPos = stFoundPos + Seperator.size();
    }
    return Lines;
}

测试代码:

std::wstring MyString(L"Part 1SEPsecond partSEPlast partSEPend");
std::vector<std::wstring> Parts = IniFile::SplitString(MyString, L"SEP");
std::wcout << L"The string: " << MyString << std::endl;
for (std::vector<std::wstring>::const_iterator it=Parts.begin(); it<Parts.end(); ++it)
{
    std::wcout << *it << L"<---" << std::endl;
}
std::wcout << std::endl;
MyString = L"this,time,a,comma separated,string";
std::wcout << L"The string: " << MyString << std::endl;
Parts = IniFile::SplitString(MyString, L",");
for (std::vector<std::wstring>::const_iterator it=Parts.begin(); it<Parts.end(); ++it)
{
    std::wcout << *it << L"<---" << std::endl;
}

测试代码的输出:

The string: Part 1SEPsecond partSEPlast partSEPend
Part 1<---
second part<---
last part<---
end<---

The string: this,time,a,comma separated,string
this<---
time<---
a<---
comma separated<---
string<---
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);
    }
}