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

这是我最喜欢的遍历字符串的方法。每个词你都可以做你想做的事。

string line = "a line of text to iterate through";
string word;

istringstream iss(line, istringstream::in);

while( iss >> word )     
{
    // Do something on `word` here...
}

其他回答

这是一个顶级答案的扩展。它现在支持设置返回元素的最大数量N。字符串的最后一位将在第N个元素中结束。MAXELEMENTS参数是可选的,如果设置为默认值0,它将返回无限数量的元素。:-)

.h:

class Myneatclass {
public:
    static std::vector<std::string>& split(const std::string &s, char delim, std::vector<std::string> &elems, const size_t MAXELEMENTS = 0);
    static std::vector<std::string> split(const std::string &s, char delim, const size_t MAXELEMENTS = 0);
};

.cpp:

std::vector<std::string>& Myneatclass::split(const std::string &s, char delim, std::vector<std::string> &elems, const size_t MAXELEMENTS) {
    std::stringstream ss(s);
    std::string item;
    while (std::getline(ss, item, delim)) {
        elems.push_back(item);
        if (MAXELEMENTS > 0 && !ss.eof() && elems.size() + 1 >= MAXELEMENTS) {
            std::getline(ss, item);
            elems.push_back(item);
            break;
        }
    }
    return elems;
}
std::vector<std::string> Myneatclass::split(const std::string &s, char delim, const size_t MAXELEMENTS) {
    std::vector<std::string> elems;
    split(s, delim, elems, MAXELEMENTS);
    return elems;
}
#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <iterator>
#include <vector>

int main() {
    using namespace std;
   int n=8;
    string sentence = "10 20 30 40 5 6 7 8";
    istringstream iss(sentence);

  vector<string> tokens;
copy(istream_iterator<string>(iss),
     istream_iterator<string>(),
     back_inserter(tokens));

     for(int i=0;i<n;i++){
        cout<<tokens.at(i);
     }


}

每个人都回答了预定义的字符串输入。我认为这个答案将帮助某人进行扫描输入。

我使用令牌向量来保存字符串令牌。这是可选的。

#include <bits/stdc++.h>

using namespace std ;
int main()
{
    string str, token ;
    getline(cin, str) ; // get the string as input
    istringstream ss(str); // insert the string into tokenizer

    vector<string> tokens; // vector tokens holds the tokens

    while (ss >> token) tokens.push_back(token); // splits the tokens
    for(auto x : tokens) cout << x << endl ; // prints the tokens

    return 0;
}


样本输入:

port city international university

样本输出:

port
city
international
university

注意,默认情况下,这将仅适用于空格作为分隔符。您可以使用自定义分隔符。为此,您定制了代码。让分隔符为“,”。所以使用

char delimiter = ',' ;
while(getline(ss, token, delimiter)) tokens.push_back(token) ;

而不是

while (ss >> token) tokens.push_back(token);
#include <iostream>
#include <string>
#include <deque>

std::deque<std::string> split(
    const std::string& line, 
    std::string::value_type delimiter,
    bool skipEmpty = false
) {
    std::deque<std::string> parts{};

    if (!skipEmpty && !line.empty() && delimiter == line.at(0)) {
        parts.push_back({});
    }

    for (const std::string::value_type& c : line) {
        if (
            (
                c == delimiter 
                &&
                (skipEmpty ? (!parts.empty() && !parts.back().empty()) : true)
            )
            ||
            (c != delimiter && parts.empty())
        ) {
            parts.push_back({});
        }

        if (c != delimiter) {
            parts.back().push_back(c);
        }
    }

    if (skipEmpty && !parts.empty() && parts.back().empty()) {
        parts.pop_back();
    }

    return parts;
}

void test(const std::string& line) {
    std::cout << line << std::endl;

    std::cout << "skipEmpty=0 |";
    for (const std::string& part : split(line, ':')) {
        std::cout << part << '|';
    }
    std::cout << std::endl;

    std::cout << "skipEmpty=1 |";
    for (const std::string& part : split(line, ':', true)) {
        std::cout << part << '|';
    }
    std::cout << std::endl;

    std::cout << std::endl;
}

int main() {
    test("foo:bar:::baz");
    test("");
    test("foo");
    test(":");
    test("::");
    test(":foo");
    test("::foo");
    test(":foo:");
    test(":foo::");

    return 0;
}

输出:

foo:bar:::baz
skipEmpty=0 |foo|bar|||baz|
skipEmpty=1 |foo|bar|baz|


skipEmpty=0 |
skipEmpty=1 |

foo
skipEmpty=0 |foo|
skipEmpty=1 |foo|

:
skipEmpty=0 |||
skipEmpty=1 |

::
skipEmpty=0 ||||
skipEmpty=1 |

:foo
skipEmpty=0 ||foo|
skipEmpty=1 |foo|

::foo
skipEmpty=0 |||foo|
skipEmpty=1 |foo|

:foo:
skipEmpty=0 ||foo||
skipEmpty=1 |foo|

:foo::
skipEmpty=0 ||foo|||
skipEmpty=1 |foo|

我知道很晚才来参加聚会,但我正在考虑最优雅的方法,如果给你一系列分隔符而不是空格,并且只使用标准库。

以下是我的想法:

要通过分隔符序列将单词拆分为字符串向量,请执行以下操作:

template<class Container>
std::vector<std::string> split_by_delimiters(const std::string& input, const Container& delimiters)
{
    std::vector<std::string> result;

    for (auto current = begin(input) ; current != end(input) ; )
    {
        auto first = find_if(current, end(input), not_in(delimiters));
        if (first == end(input)) break;
        auto last = find_if(first, end(input), is_in(delimiters));
        result.emplace_back(first, last);
        current = last;
    }
    return result;
}

通过提供一系列有效字符,以另一种方式进行拆分:

template<class Container>
std::vector<std::string> split_by_valid_chars(const std::string& input, const Container& valid_chars)
{
    std::vector<std::string> result;

    for (auto current = begin(input) ; current != end(input) ; )
    {
        auto first = find_if(current, end(input), is_in(valid_chars));
        if (first == end(input)) break;
        auto last = find_if(first, end(input), not_in(valid_chars));
        result.emplace_back(first, last);
        current = last;
    }
    return result;
}

is_in和not_in的定义如下:

namespace detail {
    template<class Container>
    struct is_in {
        is_in(const Container& charset)
        : _charset(charset)
        {}

        bool operator()(char c) const
        {
            return find(begin(_charset), end(_charset), c) != end(_charset);
        }

        const Container& _charset;
    };

    template<class Container>
    struct not_in {
        not_in(const Container& charset)
        : _charset(charset)
        {}

        bool operator()(char c) const
        {
            return find(begin(_charset), end(_charset), c) == end(_charset);
        }

        const Container& _charset;
    };

}

template<class Container>
detail::not_in<Container> not_in(const Container& c)
{
    return detail::not_in<Container>(c);
}

template<class Container>
detail::is_in<Container> is_in(const Container& c)
{
    return detail::is_in<Container>(c);
}