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

这里有一个拆分函数:

是通用的使用标准C++(无增强)接受多个分隔符忽略空标记(可以轻松更改)模板<typename T>矢量<T>拆分(常量T&str,常量T&分隔符){向量<T>v;typename T::size_type start=0;自动位置=str.find_first_of(分隔符,开始);而(pos!=T::npos){if(pos!=开始)//忽略空标记v.template_back(str,start,pos-start);开始=位置+1;pos=str.find_first_of(分隔符,开始);}if(start<str.length())//忽略尾随分隔符v.template_back(str,start,str.length()-start);//添加字符串的剩余部分返回v;}

示例用法:

    vector<string> v = split<string>("Hello, there; World", ";,");
    vector<wstring> v = split<wstring>(L"Hello, there; World", L";,");

其他回答

STL还没有这样的方法。

但是,您可以通过使用std::string::C_str()成员来使用C的strtok()函数,也可以编写自己的函数。下面是我在快速谷歌搜索(“STL字符串分割”)后找到的代码示例:

void Tokenize(const string& str,
              vector<string>& tokens,
              const string& delimiters = " ")
{
    // Skip delimiters at beginning.
    string::size_type lastPos = str.find_first_not_of(delimiters, 0);
    // Find first "non-delimiter".
    string::size_type pos     = str.find_first_of(delimiters, lastPos);

    while (string::npos != pos || string::npos != lastPos)
    {
        // Found a token, add it to the vector.
        tokens.push_back(str.substr(lastPos, pos - lastPos));
        // Skip delimiters.  Note the "not_of"
        lastPos = str.find_first_not_of(delimiters, pos);
        // Find next "non-delimiter"
        pos = str.find_first_of(delimiters, lastPos);
    }
}

摘自:http://oopweb.com/CPP/Documents/CPPHOWTO/Volume/C++编程-HOWTO-7.html

如果您对代码示例有疑问,请留下评论,我会解释。

仅仅因为它没有实现称为迭代器的typedef或重载<<运算符,并不意味着它是错误的代码。我经常使用C函数。例如,printf和scanf都比std::cin和std::cout快(很明显),fopen语法对二进制类型更友好,它们也倾向于生成更小的EXE。

不要被这种“优雅胜过性能”的交易所吸引。

有一个名为strtok的函数。

#include<string>
using namespace std;

vector<string> split(char* str,const char* delim)
{
    char* saveptr;
    char* token = strtok_r(str,delim,&saveptr);

    vector<string> result;

    while(token != NULL)
    {
        result.push_back(token);
        token = strtok_r(NULL,delim,&saveptr);
    }
    return result;
}

我使用以下代码:

namespace Core
{
    typedef std::wstring String;

    void SplitString(const Core::String& input, const Core::String& splitter, std::list<Core::String>& output)
    {
        if (splitter.empty())
        {
            throw std::invalid_argument(); // for example
        }

        std::list<Core::String> lines;

        Core::String::size_type offset = 0;

        for (;;)
        {
            Core::String::size_type splitterPos = input.find(splitter, offset);

            if (splitterPos != Core::String::npos)
            {
                lines.push_back(input.substr(offset, splitterPos - offset));
                offset = splitterPos + splitter.size();
            }
            else
            {
                lines.push_back(input.substr(offset));
                break;
            }
        }

        lines.swap(output);
    }
}

// gtest:

class SplitStringTest: public testing::Test
{
};

TEST_F(SplitStringTest, EmptyStringAndSplitter)
{
    std::list<Core::String> result;
    ASSERT_ANY_THROW(Core::SplitString(Core::String(), Core::String(), result));
}

TEST_F(SplitStringTest, NonEmptyStringAndEmptySplitter)
{
    std::list<Core::String> result;
    ASSERT_ANY_THROW(Core::SplitString(L"xy", Core::String(), result));
}

TEST_F(SplitStringTest, EmptyStringAndNonEmptySplitter)
{
    std::list<Core::String> result;
    Core::SplitString(Core::String(), Core::String(L","), result);
    ASSERT_EQ(1, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
}

TEST_F(SplitStringTest, OneCharSplitter)
{
    std::list<Core::String> result;

    Core::SplitString(L"x,y", L",", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y", *result.rbegin());

    Core::SplitString(L",xy", L",", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(L"xy", *result.rbegin());

    Core::SplitString(L"xy,", L",", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"xy", *result.begin());
    ASSERT_EQ(Core::String(), *result.rbegin());
}

TEST_F(SplitStringTest, TwoCharsSplitter)
{
    std::list<Core::String> result;

    Core::SplitString(L"x,.y,z", L",.", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y,z", *result.rbegin());

    Core::SplitString(L"x,,y,z", L",,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y,z", *result.rbegin());
}

TEST_F(SplitStringTest, RecursiveSplitter)
{
    std::list<Core::String> result;

    Core::SplitString(L",,,", L",,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(L",", *result.rbegin());

    Core::SplitString(L",.,.,", L",.,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(L".,", *result.rbegin());

    Core::SplitString(L"x,.,.,y", L",.,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L".,y", *result.rbegin());

    Core::SplitString(L",.,,.,", L",.,", result);
    ASSERT_EQ(3, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(Core::String(), *(++result.begin()));
    ASSERT_EQ(Core::String(), *result.rbegin());
}

TEST_F(SplitStringTest, NullTerminators)
{
    std::list<Core::String> result;

    Core::SplitString(L"xy", Core::String(L"\0", 1), result);
    ASSERT_EQ(1, result.size());
    ASSERT_EQ(L"xy", *result.begin());

    Core::SplitString(Core::String(L"x\0y", 3), Core::String(L"\0", 1), result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y", *result.rbegin());
}

这里有一个仅使用标准正则表达式库的正则表达式解决方案。(我有点生疏,所以可能会有一些语法错误,但这至少是一般的想法)

#include <regex.h>
#include <string.h>
#include <vector.h>

using namespace std;

vector<string> split(string s){
    regex r ("\\w+"); //regex matches whole words, (greedy, so no fragment words)
    regex_iterator<string::iterator> rit ( s.begin(), s.end(), r );
    regex_iterator<string::iterator> rend; //iterators to iterate thru words
    vector<string> result<regex_iterator>(rit, rend);
    return result;  //iterates through the matches to fill the vector
}

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;
    }
}