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

我有两条线来解决这个问题:

char sep = ' ';
std::string s="1 This is an example";

for(size_t p=0, q=0; p!=s.npos; p=q)
  std::cout << s.substr(p+(p!=0), (q=s.find(sep, p+1))-p-(p!=0)) << std::endl;

然后你可以把它放到一个向量中,而不是打印。

其他回答

#include<iostream>
#include<string>
#include<sstream>
#include<vector>
using namespace std;

    vector<string> split(const string &s, char delim) {
        vector<string> elems;
        stringstream ss(s);
        string item;
        while (getline(ss, item, delim)) {
            elems.push_back(item);
        }
        return elems;
    }

int main() {

        vector<string> x = split("thi is an sample test",' ');
        unsigned int i;
        for(i=0;i<x.size();i++)
            cout<<i<<":"<<x[i]<<endl;
        return 0;
}

我编写了以下代码。您可以指定分隔符,它可以是字符串。结果类似于Java的String.split,结果中包含空字符串。

例如,如果我们调用split(“ABCPICKABCANYABCTWO:ABC”,“ABC”),结果如下:

0  <len:0>
1 PICK <len:4>
2 ANY <len:3>
3 TWO: <len:4>
4  <len:0>

代码:

vector <string> split(const string& str, const string& delimiter = " ") {
    vector <string> tokens;

    string::size_type lastPos = 0;
    string::size_type pos = str.find(delimiter, lastPos);

    while (string::npos != pos) {
        // Found a token, add it to the vector.
        cout << str.substr(lastPos, pos - lastPos) << endl;
        tokens.push_back(str.substr(lastPos, pos - lastPos));
        lastPos = pos + delimiter.size();
        pos = str.find(delimiter, lastPos);
    }

    tokens.push_back(str.substr(lastPos, str.size() - lastPos));
    return tokens;
}

对于一个大得离谱而且可能是冗余的版本,可以尝试很多For循环。

string stringlist[10];
int count = 0;

for (int i = 0; i < sequence.length(); i++)
{
    if (sequence[i] == ' ')
    {
        stringlist[count] = sequence.substr(0, i);
        sequence.erase(0, i+1);
        i = 0;
        count++;
    }
    else if (i == sequence.length()-1)  // Last word
    {
        stringlist[count] = sequence.substr(0, i+1);
    }
}

它并不漂亮,但总的来说(除了标点符号和一系列其他错误)它是有效的!

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;
    }
}
#include <vector>
#include <string>
#include <sstream>

int main()
{
    std::string str("Split me by whitespaces");
    std::string buf;                 // Have a buffer string
    std::stringstream ss(str);       // Insert the string into a stream

    std::vector<std::string> tokens; // Create vector to hold our words

    while (ss >> buf)
        tokens.push_back(buf);

    return 0;
}