Java有一个方便的分割方法:
String str = "The quick brown fox";
String[] results = str.split(" ");
在c++中有简单的方法来做到这一点吗?
Java有一个方便的分割方法:
String str = "The quick brown fox";
String[] results = str.split(" ");
在c++中有简单的方法来做到这一点吗?
当前回答
我只是看了所有的答案,找不到下一个前提条件的解决方案:
没有动态内存分配 不使用boost 不使用正则表达式 c++17标准
这就是我的解
#include <iomanip>
#include <iostream>
#include <iterator>
#include <string_view>
#include <utility>
struct split_by_spaces
{
std::string_view text;
static constexpr char delim = ' ';
struct iterator
{
const std::string_view& text;
std::size_t cur_pos;
std::size_t end_pos;
std::string_view operator*() const
{
return { &text[cur_pos], end_pos - cur_pos };
}
bool operator==(const iterator& other) const
{
return cur_pos == other.cur_pos && end_pos == other.end_pos;
}
bool operator!=(const iterator& other) const
{
return !(*this == other);
}
iterator& operator++()
{
cur_pos = text.find_first_not_of(delim, end_pos);
if (cur_pos == std::string_view::npos)
{
cur_pos = text.size();
end_pos = cur_pos;
return *this;
}
end_pos = text.find(delim, cur_pos);
if (cur_pos == std::string_view::npos)
{
end_pos = text.size();
}
return *this;
}
};
[[nodiscard]] iterator begin() const
{
auto start = text.find_first_not_of(delim);
if (start == std::string_view::npos)
{
return iterator{ text, text.size(), text.size() };
}
auto end_word = text.find(delim, start);
if (end_word == std::string_view::npos)
{
end_word = text.size();
}
return iterator{ text, start, end_word };
}
[[nodiscard]] iterator end() const
{
return iterator{ text, text.size(), text.size() };
}
};
int main(int argc, char** argv)
{
using namespace std::literals;
auto str = " there should be no memory allocation during parsing"
" into words this line and you should'n create any"
" contaner for intermediate words "sv;
auto comma = "";
for (std::string_view word : split_by_spaces{ str })
{
std::cout << std::exchange(comma, ",") << std::quoted(word);
}
auto only_spaces = " "sv;
for (std::string_view word : split_by_spaces{ only_spaces })
{
std::cout << "you will not see this line in output" << std::endl;
}
}
其他回答
使用regex_token_iterators的解决方案:
#include <iostream>
#include <regex>
#include <string>
using namespace std;
int main()
{
string str("The quick brown fox");
regex reg("\\s+");
sregex_token_iterator iter(str.begin(), str.end(), reg, -1);
sregex_token_iterator end;
vector<string> vec(iter, end);
for (auto a : vec)
{
cout << a << endl;
}
}
请看这个例子。它可能对你有帮助。
#include <iostream>
#include <sstream>
using namespace std;
int main ()
{
string tmps;
istringstream is ("the dellimiter is the space");
while (is.good ()) {
is >> tmps;
cout << tmps << "\n";
}
return 0;
}
下面是我的Swiss®军刀字符串标记器,用于用空格分隔字符串,处理单引号和双引号包装的字符串,以及从结果中剥离这些字符。我使用RegexBuddy 4。x生成大部分代码片段,但我添加了用于剥离引号和其他一些东西的自定义处理。
#include <string>
#include <locale>
#include <regex>
std::vector<std::wstring> tokenize_string(std::wstring string_to_tokenize) {
std::vector<std::wstring> tokens;
std::wregex re(LR"(("[^"]*"|'[^']*'|[^"' ]+))", std::regex_constants::collate);
std::wsregex_iterator next( string_to_tokenize.begin(),
string_to_tokenize.end(),
re,
std::regex_constants::match_not_null );
std::wsregex_iterator end;
const wchar_t single_quote = L'\'';
const wchar_t double_quote = L'\"';
while ( next != end ) {
std::wsmatch match = *next;
const std::wstring token = match.str( 0 );
next++;
if (token.length() > 2 && (token.front() == double_quote || token.front() == single_quote))
tokens.emplace_back( std::wstring(token.begin()+1, token.begin()+token.length()-1) );
else
tokens.emplace_back(token);
}
return tokens;
}
另一种快速方法是使用getline。喜欢的东西:
stringstream ss("bla bla");
string s;
while (getline(ss, s, ' ')) {
cout << s << endl;
}
如果需要,可以创建一个简单的split()方法,返回vector<string>,即 真的有用。
您可以使用流、迭代器和复制算法来相当直接地做到这一点。
#include <string>
#include <vector>
#include <iostream>
#include <istream>
#include <ostream>
#include <iterator>
#include <sstream>
#include <algorithm>
int main()
{
std::string str = "The quick brown fox";
// construct a stream from the string
std::stringstream strstr(str);
// use stream iterators to copy the stream to the vector as whitespace separated strings
std::istream_iterator<std::string> it(strstr);
std::istream_iterator<std::string> end;
std::vector<std::string> results(it, end);
// send the vector to stdout.
std::ostream_iterator<std::string> oit(std::cout);
std::copy(results.begin(), results.end(), oit);
}