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
这类似于堆栈溢出问题：如何在C++中标记字符串？。需要Boost外部库
#include <iostream>
#include <string>
#include <boost/tokenizer.hpp>
using namespace std;
using namespace boost;
int main(int argc, char** argv)
{
string text = "token test\tstring";
char_separator<char> sep(" \t");
tokenizer<char_separator<char>> tokens(text, sep);
for (const string& t : tokens)
{
cout << t << "." << endl;
}
}
其他回答
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！
使用std:：string_view和Eric Niebler的range-v3库：
https://wandbox.org/permlink/kW5lwRCL1pxjp2pW
#include <iostream>
#include <string>
#include <string_view>
#include "range/v3/view.hpp"
#include "range/v3/algorithm.hpp"
int main() {
std::string s = "Somewhere down the range v3 library";
ranges::for_each(s
| ranges::view::split(' ')
| ranges::view::transform([](auto &&sub) {
return std::string_view(&*sub.begin(), ranges::distance(sub));
}),
[](auto s) {std::cout << "Substring: " << s << "\n";}
);
}
通过使用循环的范围而不是范围：：for_each算法：
#include <iostream>
#include <string>
#include <string_view>
#include "range/v3/view.hpp"
int main()
{
std::string str = "Somewhere down the range v3 library";
for (auto s : str | ranges::view::split(' ')
| ranges::view::transform([](auto&& sub) { return std::string_view(&*sub.begin(), ranges::distance(sub)); }
))
{
std::cout << "Substring: " << s << "\n";
}
}
我的代码是：
#include <list>
#include <string>
template<class StringType = std::string, class ContainerType = std::list<StringType> >
class DSplitString:public ContainerType
{
public:
explicit DSplitString(const StringType& strString, char cChar, bool bSkipEmptyParts = true)
{
size_t iPos = 0;
size_t iPos_char = 0;
while(StringType::npos != (iPos_char = strString.find(cChar, iPos)))
{
StringType strTemp = strString.substr(iPos, iPos_char - iPos);
if((bSkipEmptyParts && !strTemp.empty()) || (!bSkipEmptyParts))
push_back(strTemp);
iPos = iPos_char + 1;
}
}
explicit DSplitString(const StringType& strString, const StringType& strSub, bool bSkipEmptyParts = true)
{
size_t iPos = 0;
size_t iPos_char = 0;
while(StringType::npos != (iPos_char = strString.find(strSub, iPos)))
{
StringType strTemp = strString.substr(iPos, iPos_char - iPos);
if((bSkipEmptyParts && !strTemp.empty()) || (!bSkipEmptyParts))
push_back(strTemp);
iPos = iPos_char + strSub.length();
}
}
};
例子：
#include <iostream>
#include <string>
int _tmain(int argc, _TCHAR* argv[])
{
DSplitString<> aa("doicanhden1;doicanhden2;doicanhden3;", ';');
for each (std::string var in aa)
{
std::cout << var << std::endl;
}
std::cin.get();
return 0;
}
短而优雅
#include <vector>
#include <string>
using namespace std;
vector<string> split(string data, string token)
{
vector<string> output;
size_t pos = string::npos; // size_t to avoid improbable overflow
do
{
pos = data.find(token);
output.push_back(data.substr(0, pos));
if (string::npos != pos)
data = data.substr(pos + token.size());
} while (string::npos != pos);
return output;
}
可以使用任何字符串作为分隔符，也可以与二进制数据一起使用（std:：string支持二进制数据，包括空值）
使用：
auto a = split("this!!is!!!example!string", "!!");
输出：
this
is
!example!string
对于那个些需要使用字符串分隔符拆分字符串的人，也许可以尝试我的以下解决方案。
std::vector<size_t> str_pos(const std::string &search, const std::string &target)
{
std::vector<size_t> founds;
if(!search.empty())
{
size_t start_pos = 0;
while (true)
{
size_t found_pos = target.find(search, start_pos);
if(found_pos != std::string::npos)
{
size_t found = found_pos;
founds.push_back(found);
start_pos = (found_pos + 1);
}
else
{
break;
}
}
}
return founds;
}
std::string str_sub_index(size_t begin_index, size_t end_index, const std::string &target)
{
std::string sub;
size_t size = target.length();
const char* copy = target.c_str();
for(size_t i = begin_index; i <= end_index; i++)
{
if(i >= size)
{
break;
}
else
{
char c = copy[i];
sub += c;
}
}
return sub;
}
std::vector<std::string> str_split(const std::string &delimiter, const std::string &target)
{
std::vector<std::string> splits;
if(!delimiter.empty())
{
std::vector<size_t> founds = str_pos(delimiter, target);
size_t founds_size = founds.size();
if(founds_size > 0)
{
size_t search_len = delimiter.length();
size_t begin_index = 0;
for(int i = 0; i <= founds_size; i++)
{
std::string sub;
if(i != founds_size)
{
size_t pos = founds.at(i);
sub = str_sub_index(begin_index, pos - 1, target);
begin_index = (pos + search_len);
}
else
{
sub = str_sub_index(begin_index, (target.length() - 1), target);
}
splits.push_back(sub);
}
}
}
return splits;
}
这些片段由3个函数组成。坏消息是使用str_split函数，您将需要另外两个函数。是的，这是一大块代码。但好消息是，这两个附加功能可以独立工作，有时也很有用
测试main（）块中的函数如下：
int main()
{
std::string s = "Hello, world! We need to make the world a better place. Because your world is also my world, and our children's world.";
std::vector<std::string> split = str_split("world", s);
for(int i = 0; i < split.size(); i++)
{
std::cout << split[i] << std::endl;
}
}
它将产生：
Hello,
! We need to make the
a better place. Because your
is also my
, and our children's
.
我认为这不是最有效的代码，但至少它可以工作。希望有帮助。
