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

谢谢@Jairo Abdiel Toribio Cisneros。它对我有效,但您的函数返回一些空元素。因此,对于没有空的返回,我编辑了以下内容:

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

    for(std::string::const_iterator i = str.begin(); i <= str.end(); ++i) {
        if(*i != *delim && i != str.end()) {
            tmp += *i;
        } else {
            if (tmp.length() > 0) {
                v.push_back(tmp);
            }
            tmp = "";
        }
    }

    return v;
}

使用:

std::string s = "one:two::three";
std::string delim = ":";
std::vector<std::string> vv = split(s, delim.c_str());

其他回答

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...
}

我有一种与其他解决方案非常不同的方法,它提供了很多其他解决方案所缺乏的价值,但当然也有其缺点。这是一个工作实现,示例是在单词周围放置<tag></tag>。

首先,这个问题可以通过一个循环解决,不需要额外的内存,只需考虑四种逻辑情况。从概念上讲,我们对边界感兴趣。我们的代码应该反映出这一点:让我们遍历字符串,一次查看两个字符,记住字符串的开头和结尾都有特殊情况。

缺点是我们必须编写实现,这有点冗长,但大多是方便的样板。

好处是我们编写了实现,因此很容易根据特定的需要定制它,例如区分左和写单词边界,使用任何一组分隔符,或处理其他情况,例如无边界或错误位置。

using namespace std;

#include <iostream>
#include <string>

#include <cctype>

typedef enum boundary_type_e {
    E_BOUNDARY_TYPE_ERROR = -1,
    E_BOUNDARY_TYPE_NONE,
    E_BOUNDARY_TYPE_LEFT,
    E_BOUNDARY_TYPE_RIGHT,
} boundary_type_t;

typedef struct boundary_s {
    boundary_type_t type;
    int pos;
} boundary_t;

bool is_delim_char(int c) {
    return isspace(c); // also compare against any other chars you want to use as delimiters
}

bool is_word_char(int c) {
    return ' ' <= c && c <= '~' && !is_delim_char(c);
}

boundary_t maybe_word_boundary(string str, int pos) {
    int len = str.length();
    if (pos < 0 || pos >= len) {
        return (boundary_t){.type = E_BOUNDARY_TYPE_ERROR};
    } else {
        if (pos == 0 && is_word_char(str[pos])) {
            // if the first character is word-y, we have a left boundary at the beginning
            return (boundary_t){.type = E_BOUNDARY_TYPE_LEFT, .pos = pos};
        } else if (pos == len - 1 && is_word_char(str[pos])) {
            // if the last character is word-y, we have a right boundary left of the null terminator
            return (boundary_t){.type = E_BOUNDARY_TYPE_RIGHT, .pos = pos + 1};
        } else if (!is_word_char(str[pos]) && is_word_char(str[pos + 1])) {
            // if we have a delimiter followed by a word char, we have a left boundary left of the word char
            return (boundary_t){.type = E_BOUNDARY_TYPE_LEFT, .pos = pos + 1};
        } else if (is_word_char(str[pos]) && !is_word_char(str[pos + 1])) {
            // if we have a word char followed by a delimiter, we have a right boundary right of the word char
            return (boundary_t){.type = E_BOUNDARY_TYPE_RIGHT, .pos = pos + 1};
        }
        return (boundary_t){.type = E_BOUNDARY_TYPE_NONE};
    }
}

int main() {
    string str;
    getline(cin, str);

    int len = str.length();
    for (int i = 0; i < len; i++) {
        boundary_t boundary = maybe_word_boundary(str, i);
        if (boundary.type == E_BOUNDARY_TYPE_LEFT) {
            // whatever
        } else if (boundary.type == E_BOUNDARY_TYPE_RIGHT) {
            // whatever
        }
    }
}

正如您所看到的,代码非常容易理解和微调,代码的实际使用非常简短和简单。使用C++不应阻止我们编写最简单、最容易定制的代码,即使这意味着不使用STL。我认为这是Linus Torvalds所说的“品味”的一个例子,因为我们已经消除了所有不需要的逻辑,而写作风格自然允许在需要处理的时候处理更多的案件。

可以改进此代码的可能是使用enum类,在maybe_word_boundary中接受指向is_word_char的函数指针,而不是直接调用is_word_char,并传递lambda。

根据Galik的回答,我做了这个。这大部分都在这里,所以我不必一遍又一遍地写。C++仍然没有原生拆分函数,这真是太疯狂了。特征:

应该很快。容易理解(我认为)。合并空节。使用多个分隔符(例如“\r\n”)很简单

#include <string>
#include <vector>
#include <algorithm>

std::vector<std::string> split(const std::string& s, const std::string& delims)
{
    using namespace std;

    vector<string> v;

    // Start of an element.
    size_t elemStart = 0;

    // We start searching from the end of the previous element, which
    // initially is the start of the string.
    size_t elemEnd = 0;

    // Find the first non-delim, i.e. the start of an element, after the end of the previous element.
    while((elemStart = s.find_first_not_of(delims, elemEnd)) != string::npos)
    {
        // Find the first delem, i.e. the end of the element (or if this fails it is the end of the string).
        elemEnd = s.find_first_of(delims, elemStart);
        // Add it.
        v.emplace_back(s, elemStart, elemEnd == string::npos ? string::npos : elemEnd - elemStart);
    }
    // When there are no more non-spaces, we are done.

    return v;
}

一些C++20编译器和大多数C++23编译器(range和string_view)

for (auto word : std::views::split("Somewhere down the road", ' '))
        std::cout << std::string_view{ word.begin(), word.end() } << std::endl;

使用vector作为基类的快速版本,可完全访问其所有运算符:

    // Split string into parts.
    class Split : public std::vector<std::string>
    {
        public:
            Split(const std::string& str, char* delimList)
            {
               size_t lastPos = 0;
               size_t pos = str.find_first_of(delimList);

               while (pos != std::string::npos)
               {
                    if (pos != lastPos)
                        push_back(str.substr(lastPos, pos-lastPos));
                    lastPos = pos + 1;
                    pos = str.find_first_of(delimList, lastPos);
               }
               if (lastPos < str.length())
                   push_back(str.substr(lastPos, pos-lastPos));
            }
    };

用于填充STL集的示例:

std::set<std::string> words;
Split split("Hello,World", ",");
words.insert(split.begin(), split.end());