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
在我看来很奇怪的是，SO网站上有这么多注重速度的书呆子，却没有人给出一个使用编译时生成的分隔符查找表的版本(下面是示例实现)。使用查找表和迭代器应该在效率上击败std::regex，如果你不需要击败regex，就使用它，它是c++ 11的标准，超级灵活。
有些人已经建议使用正则表达式，但对于新手来说，这里有一个打包的示例，应该完全符合OP的期望:
std::vector<std::string> split(std::string::const_iterator it, std::string::const_iterator end, std::regex e = std::regex{"\\w+"}){
std::smatch m{};
std::vector<std::string> ret{};
while (std::regex_search (it,end,m,e)) {
ret.emplace_back(m.str());
std::advance(it, m.position() + m.length()); //next start position = match position + match length
}
return ret;
}
std::vector<std::string> split(const std::string &s, std::regex e = std::regex{"\\w+"}){ //comfort version calls flexible version
return split(s.cbegin(), s.cend(), std::move(e));
}
int main ()
{
std::string str {"Some people, excluding those present, have been compile time constants - since puberty."};
auto v = split(str);
for(const auto&s:v){
std::cout << s << std::endl;
}
std::cout << "crazy version:" << std::endl;
v = split(str, std::regex{"[^e]+"}); //using e as delim shows flexibility
for(const auto&s:v){
std::cout << s << std::endl;
}
return 0;
}
如果我们需要更快并接受所有字符必须为8位的约束，我们可以在编译时使用元编程创建一个查找表:
template<bool...> struct BoolSequence{}; //just here to hold bools
template<char...> struct CharSequence{}; //just here to hold chars
template<typename T, char C> struct Contains; //generic
template<char First, char... Cs, char Match> //not first specialization
struct Contains<CharSequence<First, Cs...>,Match> :
Contains<CharSequence<Cs...>, Match>{}; //strip first and increase index
template<char First, char... Cs> //is first specialization
struct Contains<CharSequence<First, Cs...>,First>: std::true_type {};
template<char Match> //not found specialization
struct Contains<CharSequence<>,Match>: std::false_type{};
template<int I, typename T, typename U>
struct MakeSequence; //generic
template<int I, bool... Bs, typename U>
struct MakeSequence<I,BoolSequence<Bs...>, U>: //not last
MakeSequence<I-1, BoolSequence<Contains<U,I-1>::value,Bs...>, U>{};
template<bool... Bs, typename U>
struct MakeSequence<0,BoolSequence<Bs...>,U>{ //last
using Type = BoolSequence<Bs...>;
};
template<typename T> struct BoolASCIITable;
template<bool... Bs> struct BoolASCIITable<BoolSequence<Bs...>>{
/* could be made constexpr but not yet supported by MSVC */
static bool isDelim(const char c){
static const bool table[256] = {Bs...};
return table[static_cast<int>(c)];
}
};
using Delims = CharSequence<'.',',',' ',':','\n'>; //list your custom delimiters here
using Table = BoolASCIITable<typename MakeSequence<256,BoolSequence<>,Delims>::Type>;
有了这些，创建getNextToken函数就很容易了:
template<typename T_It>
std::pair<T_It,T_It> getNextToken(T_It begin,T_It end){
begin = std::find_if(begin,end,std::not1(Table{})); //find first non delim or end
auto second = std::find_if(begin,end,Table{}); //find first delim or end
return std::make_pair(begin,second);
}
使用它也很简单:
int main() {
std::string s{"Some people, excluding those present, have been compile time constants - since puberty."};
auto it = std::begin(s);
auto end = std::end(s);
while(it != std::end(s)){
auto token = getNextToken(it,end);
std::cout << std::string(token.first,token.second) << std::endl;
it = token.second;
}
return 0;
}
这里有一个生动的例子:http://ideone.com/GKtkLQ
其他回答
简单的c++代码(标准c++ 98)，接受多个分隔符(在std::string中指定)，只使用向量、字符串和迭代器。
#include <iostream>
#include <vector>
#include <string>
#include <stdexcept>
std::vector<std::string>
split(const std::string& str, const std::string& delim){
std::vector<std::string> result;
if (str.empty())
throw std::runtime_error("Can not tokenize an empty string!");
std::string::const_iterator begin, str_it;
begin = str_it = str.begin();
do {
while (delim.find(*str_it) == std::string::npos && str_it != str.end())
str_it++; // find the position of the first delimiter in str
std::string token = std::string(begin, str_it); // grab the token
if (!token.empty()) // empty token only when str starts with a delimiter
result.push_back(token); // push the token into a vector<string>
while (delim.find(*str_it) != std::string::npos && str_it != str.end())
str_it++; // ignore the additional consecutive delimiters
begin = str_it; // process the remaining tokens
} while (str_it != str.end());
return result;
}
int main() {
std::string test_string = ".this is.a.../.simple;;test;;;END";
std::string delim = "; ./"; // string containing the delimiters
std::vector<std::string> tokens = split(test_string, delim);
for (std::vector<std::string>::const_iterator it = tokens.begin();
it != tokens.end(); it++)
std::cout << *it << std::endl;
}
我为自己编写了一个https://stackoverflow.com/a/50247503/3976739的简化版本(可能有一点效率)。我希望这能有所帮助。
void StrTokenizer(string& source, const char* delimiter, vector<string>& Tokens)
{
size_t new_index = 0;
size_t old_index = 0;
while (new_index != std::string::npos)
{
new_index = source.find(delimiter, old_index);
Tokens.emplace_back(source.substr(old_index, new_index-old_index));
if (new_index != std::string::npos)
old_index = ++new_index;
}
}
在我看来很奇怪的是，SO网站上有这么多注重速度的书呆子，却没有人给出一个使用编译时生成的分隔符查找表的版本(下面是示例实现)。使用查找表和迭代器应该在效率上击败std::regex，如果你不需要击败regex，就使用它，它是c++ 11的标准，超级灵活。
有些人已经建议使用正则表达式，但对于新手来说，这里有一个打包的示例，应该完全符合OP的期望:
std::vector<std::string> split(std::string::const_iterator it, std::string::const_iterator end, std::regex e = std::regex{"\\w+"}){
std::smatch m{};
std::vector<std::string> ret{};
while (std::regex_search (it,end,m,e)) {
ret.emplace_back(m.str());
std::advance(it, m.position() + m.length()); //next start position = match position + match length
}
return ret;
}
std::vector<std::string> split(const std::string &s, std::regex e = std::regex{"\\w+"}){ //comfort version calls flexible version
return split(s.cbegin(), s.cend(), std::move(e));
}
int main ()
{
std::string str {"Some people, excluding those present, have been compile time constants - since puberty."};
auto v = split(str);
for(const auto&s:v){
std::cout << s << std::endl;
}
std::cout << "crazy version:" << std::endl;
v = split(str, std::regex{"[^e]+"}); //using e as delim shows flexibility
for(const auto&s:v){
std::cout << s << std::endl;
}
return 0;
}
如果我们需要更快并接受所有字符必须为8位的约束，我们可以在编译时使用元编程创建一个查找表:
template<bool...> struct BoolSequence{}; //just here to hold bools
template<char...> struct CharSequence{}; //just here to hold chars
template<typename T, char C> struct Contains; //generic
template<char First, char... Cs, char Match> //not first specialization
struct Contains<CharSequence<First, Cs...>,Match> :
Contains<CharSequence<Cs...>, Match>{}; //strip first and increase index
template<char First, char... Cs> //is first specialization
struct Contains<CharSequence<First, Cs...>,First>: std::true_type {};
template<char Match> //not found specialization
struct Contains<CharSequence<>,Match>: std::false_type{};
template<int I, typename T, typename U>
struct MakeSequence; //generic
template<int I, bool... Bs, typename U>
struct MakeSequence<I,BoolSequence<Bs...>, U>: //not last
MakeSequence<I-1, BoolSequence<Contains<U,I-1>::value,Bs...>, U>{};
template<bool... Bs, typename U>
struct MakeSequence<0,BoolSequence<Bs...>,U>{ //last
using Type = BoolSequence<Bs...>;
};
template<typename T> struct BoolASCIITable;
template<bool... Bs> struct BoolASCIITable<BoolSequence<Bs...>>{
/* could be made constexpr but not yet supported by MSVC */
static bool isDelim(const char c){
static const bool table[256] = {Bs...};
return table[static_cast<int>(c)];
}
};
using Delims = CharSequence<'.',',',' ',':','\n'>; //list your custom delimiters here
using Table = BoolASCIITable<typename MakeSequence<256,BoolSequence<>,Delims>::Type>;
有了这些，创建getNextToken函数就很容易了:
template<typename T_It>
std::pair<T_It,T_It> getNextToken(T_It begin,T_It end){
begin = std::find_if(begin,end,std::not1(Table{})); //find first non delim or end
auto second = std::find_if(begin,end,Table{}); //find first delim or end
return std::make_pair(begin,second);
}
使用它也很简单:
int main() {
std::string s{"Some people, excluding those present, have been compile time constants - since puberty."};
auto it = std::begin(s);
auto end = std::end(s);
while(it != std::end(s)){
auto token = getNextToken(it,end);
std::cout << std::string(token.first,token.second) << std::endl;
it = token.second;
}
return 0;
}
这里有一个生动的例子:http://ideone.com/GKtkLQ
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;
}
Adam Pierce的回答提供了一个采用const char*的手工标记器。使用迭代器会有一些问题，因为对字符串的结束迭代器进行递增是未定义的。也就是说，给定字符串str{"The quick brown fox"}，我们当然可以做到:
auto start = find(cbegin(str), cend(str), ' ');
vector<string> tokens{ string(cbegin(str), start) };
while (start != cend(str)) {
const auto finish = find(++start, cend(str), ' ');
tokens.push_back(string(start, finish));
start = finish;
}
生活的例子
如果你想通过使用标准功能来抽象复杂性，On Freund建议strtok是一个简单的选择:
vector<string> tokens;
for (auto i = strtok(data(str), " "); i != nullptr; i = strtok(nullptr, " ")) tokens.push_back(i);
如果你不能访问c++ 17，你需要像这个例子一样替换data(str): http://ideone.com/8kAGoa
虽然在示例中没有演示，但strtok不需要为每个标记使用相同的分隔符。除了这个优势，还有几个缺点:
strtok cannot be used on multiple strings at the same time: Either a nullptr must be passed to continue tokenizing the current string or a new char* to tokenize must be passed (there are some non-standard implementations which do support this however, such as: strtok_s) For the same reason strtok cannot be used on multiple threads simultaneously (this may however be implementation defined, for example: Visual Studio's implementation is thread safe) Calling strtok modifies the string it is operating on, so it cannot be used on const strings, const char*s, or literal strings, to tokenize any of these with strtok or to operate on a string who's contents need to be preserved, str would have to be copied, then the copy could be operated on
c++20为我们提供了split_view来以非破坏性的方式标记字符串:https://topanswers.xyz/cplusplus?q=749#a874
前面的方法不能就地生成标记化的向量，这意味着如果不将它们抽象为辅助函数，它们就不能初始化const vector<string>令牌。该功能和接受任何空白分隔符的能力可以使用istream_iterator来利用。例如，给定const string str{"The quick \tbrown \nfox"}，我们可以这样做:
istringstream is{ str };
const vector<string> tokens{ istream_iterator<string>(is), istream_iterator<string>() };
生活的例子
对于这个选项，需要构造一个istringstream的代价比前面两个选项要大得多，但是这个代价通常隐藏在字符串分配的代价中。
如果上面的选项都不够灵活，不能满足您的标记化需求，那么最灵活的选项是使用regex_token_iterator，当然这种灵活性会带来更大的开销，但同样，这可能隐藏在字符串分配成本中。例如，我们想要基于非转义的逗号进行标记化，也吃空白，给定以下输入:const string str{" the,qu\\，ick，\tbrown, fox"}我们可以这样做:
const regex re{ "\\s*((?:[^\\\\,]|\\\\.)*?)\\s*(?:,|$)" };
const vector<string> tokens{ sregex_token_iterator(cbegin(str), cend(str), re, 1), sregex_token_iterator() };
生活的例子
