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
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;
其他回答
使用Boost的可能解决方案可能是：
#include <boost/algorithm/string.hpp>
std::vector<std::string> strs;
boost::split(strs, "string to split", boost::is_any_of("\t "));
这种方法可能比字符串流方法更快。由于这是一个通用模板函数，因此可以使用各种分隔符拆分其他类型的字符串（wchar等或UTF-8）。
有关详细信息，请参阅文档。
并不是说我们需要更多的答案，但这是我受到埃文·特兰启发后想到的。
std::vector <std::string> split(const string &input, auto delimiter, bool skipEmpty=true) {
/*
Splits a string at each delimiter and returns these strings as a string vector.
If the delimiter is not found then nothing is returned.
If skipEmpty is true then strings between delimiters that are 0 in length will be skipped.
*/
bool delimiterFound = false;
int pos=0, pPos=0;
std::vector <std::string> result;
while (true) {
pos = input.find(delimiter,pPos);
if (pos != std::string::npos) {
if (skipEmpty==false or pos-pPos > 0) // if empty values are to be kept or not
result.push_back(input.substr(pPos,pos-pPos));
delimiterFound = true;
} else {
if (pPos < input.length() and delimiterFound) {
if (skipEmpty==false or input.length()-pPos > 0) // if empty values are to be kept or not
result.push_back(input.substr(pPos,input.length()-pPos));
}
break;
}
pPos = pos+1;
}
return result;
}
我使用以下方法
void split(string in, vector<string>& parts, char separator) {
string::iterator ts, curr;
ts = curr = in.begin();
for(; curr <= in.end(); curr++ ) {
if( (curr == in.end() || *curr == separator) && curr > ts )
parts.push_back( string( ts, curr ));
if( curr == in.end() )
break;
if( *curr == separator ) ts = curr + 1;
}
}
PlasmaHH，我忘记包含删除带有空格的标记的额外检查（curr>ts）。
这是我写的一个函数，帮助我做了很多事情。它在为WebSocket做协议时帮助了我。
using namespace std;
#include <iostream>
#include <vector>
#include <sstream>
#include <string>
vector<string> split ( string input , string split_id ) {
vector<string> result;
int i = 0;
bool add;
string temp;
stringstream ss;
size_t found;
string real;
int r = 0;
while ( i != input.length() ) {
add = false;
ss << input.at(i);
temp = ss.str();
found = temp.find(split_id);
if ( found != string::npos ) {
add = true;
real.append ( temp , 0 , found );
} else if ( r > 0 && ( i+1 ) == input.length() ) {
add = true;
real.append ( temp , 0 , found );
}
if ( add ) {
result.push_back(real);
ss.str(string());
ss.clear();
temp.clear();
real.clear();
r = 0;
}
i++;
r++;
}
return result;
}
int main() {
string s = "S,o,m,e,w,h,e,r,e, down the road \n In a really big C++ house. \n Lives a little old lady. \n That no one ever knew. \n She comes outside. \n In the very hot sun. \n\n\n\n\n\n\n\n And throws C++ at us. \n The End. FIN.";
vector < string > Token;
Token = split ( s , "," );
for ( int i = 0 ; i < Token.size(); i++) cout << Token.at(i) << endl;
cout << endl << Token.size();
int a;
cin >> a;
return a;
}
如果您需要通过非空格符号解析字符串，则字符串流可能很方便：
string s = "Name:JAck; Spouse:Susan; ...";
string dummy, name, spouse;
istringstream iss(s);
getline(iss, dummy, ':');
getline(iss, name, ';');
getline(iss, dummy, ':');
getline(iss, spouse, ';')
