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);
}
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;
}
谢谢@Jairo Abdiel Toribio Cisneros。它对我有效,但您的函数返回一些空元素。因此,对于没有空的返回,我编辑了以下内容:
std::vector<std::string> split(std::string str, const char* delim) {
std::vector<std::string> v;
std::string tmp;
for(std::string::const_iterator i = str.begin(); i <= str.end(); ++i) {
if(*i != *delim && i != str.end()) {
tmp += *i;
} else {
if (tmp.length() > 0) {
v.push_back(tmp);
}
tmp = "";
}
}
return v;
}
使用:
std::string s = "one:two::three";
std::string delim = ":";
std::vector<std::string> vv = split(s, delim.c_str());
STL还没有这样的方法。
但是,您可以通过使用std::string::C_str()成员来使用C的strtok()函数,也可以编写自己的函数。下面是我在快速谷歌搜索(“STL字符串分割”)后找到的代码示例:
void Tokenize(const string& str,
vector<string>& tokens,
const string& delimiters = " ")
{
// Skip delimiters at beginning.
string::size_type lastPos = str.find_first_not_of(delimiters, 0);
// Find first "non-delimiter".
string::size_type pos = str.find_first_of(delimiters, lastPos);
while (string::npos != pos || string::npos != lastPos)
{
// Found a token, add it to the vector.
tokens.push_back(str.substr(lastPos, pos - lastPos));
// Skip delimiters. Note the "not_of"
lastPos = str.find_first_not_of(delimiters, pos);
// Find next "non-delimiter"
pos = str.find_first_of(delimiters, lastPos);
}
}
摘自:http://oopweb.com/CPP/Documents/CPPHOWTO/Volume/C++编程-HOWTO-7.html
如果您对代码示例有疑问,请留下评论,我会解释。
仅仅因为它没有实现称为迭代器的typedef或重载<<运算符,并不意味着它是错误的代码。我经常使用C函数。例如,printf和scanf都比std::cin和std::cout快(很明显),fopen语法对二进制类型更友好,它们也倾向于生成更小的EXE。
不要被这种“优雅胜过性能”的交易所吸引。
下面是一个更好的方法。它可以接受任何字符,除非您愿意,否则不会拆分行。不需要特殊的库(嗯,除了std,但谁真的认为这是一个额外的库),没有指针,没有引用,而且它是静态的。只是简单的C++。
#pragma once
#include <vector>
#include <sstream>
using namespace std;
class Helpers
{
public:
static vector<string> split(string s, char delim)
{
stringstream temp (stringstream::in | stringstream::out);
vector<string> elems(0);
if (s.size() == 0 || delim == 0)
return elems;
for(char c : s)
{
if(c == delim)
{
elems.push_back(temp.str());
temp = stringstream(stringstream::in | stringstream::out);
}
else
temp << c;
}
if (temp.str().size() > 0)
elems.push_back(temp.str());
return elems;
}
//Splits string s with a list of delimiters in delims (it's just a list, like if we wanted to
//split at the following letters, a, b, c we would make delims="abc".
static vector<string> split(string s, string delims)
{
stringstream temp (stringstream::in | stringstream::out);
vector<string> elems(0);
bool found;
if(s.size() == 0 || delims.size() == 0)
return elems;
for(char c : s)
{
found = false;
for(char d : delims)
{
if (c == d)
{
elems.push_back(temp.str());
temp = stringstream(stringstream::in | stringstream::out);
found = true;
break;
}
}
if(!found)
temp << c;
}
if(temp.str().size() > 0)
elems.push_back(temp.str());
return elems;
}
};