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);
}
仅为方便:
template<class V, typename T>
bool in(const V &v, const T &el) {
return std::find(v.begin(), v.end(), el) != v.end();
}
基于多个分隔符的实际拆分:
std::vector<std::string> split(const std::string &s,
const std::vector<char> &delims) {
std::vector<std::string> res;
auto stuff = [&delims](char c) { return !in(delims, c); };
auto space = [&delims](char c) { return in(delims, c); };
auto first = std::find_if(s.begin(), s.end(), stuff);
while (first != s.end()) {
auto last = std::find_if(first, s.end(), space);
res.push_back(std::string(first, last));
first = std::find_if(last + 1, s.end(), stuff);
}
return res;
}
用法:
int main() {
std::string s = " aaa, bb cc ";
for (auto el: split(s, {' ', ','}))
std::cout << el << std::endl;
return 0;
}
我相信还没有人发布这个解决方案。与其直接使用分隔符,它基本上与boost::split()相同,即它允许您传递一个谓词,如果字符是分隔符,则返回true,否则返回false。我认为这给了程序员更多的控制,最棒的是你不需要提升。
template <class Container, class String, class Predicate>
void split(Container& output, const String& input,
const Predicate& pred, bool trimEmpty = false) {
auto it = begin(input);
auto itLast = it;
while (it = find_if(it, end(input), pred), it != end(input)) {
if (not (trimEmpty and it == itLast)) {
output.emplace_back(itLast, it);
}
++it;
itLast = it;
}
}
然后可以这样使用:
struct Delim {
bool operator()(char c) {
return not isalpha(c);
}
};
int main() {
string s("#include<iostream>\n"
"int main() { std::cout << \"Hello world!\" << std::endl; }");
vector<string> v;
split(v, s, Delim(), true);
/* Which is also the same as */
split(v, s, [](char c) { return not isalpha(c); }, true);
for (const auto& i : v) {
cout << i << endl;
}
}
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;
}
};
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;
}
这是我的版本获取了Kev的来源:
#include <string>
#include <vector>
void split(vector<string> &result, string str, char delim ) {
string tmp;
string::iterator i;
result.clear();
for(i = str.begin(); i <= str.end(); ++i) {
if((const char)*i != delim && i != str.end()) {
tmp += *i;
} else {
result.push_back(tmp);
tmp = "";
}
}
}
之后,调用函数并执行以下操作:
vector<string> hosts;
split(hosts, "192.168.1.2,192.168.1.3", ',');
for( size_t i = 0; i < hosts.size(); i++){
cout << "Connecting host : " << hosts.at(i) << "..." << endl;
}