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
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;
}
}
其他回答
#include <iostream>
#include <string>
#include <deque>
std::deque<std::string> split(
const std::string& line,
std::string::value_type delimiter,
bool skipEmpty = false
) {
std::deque<std::string> parts{};
if (!skipEmpty && !line.empty() && delimiter == line.at(0)) {
parts.push_back({});
}
for (const std::string::value_type& c : line) {
if (
(
c == delimiter
&&
(skipEmpty ? (!parts.empty() && !parts.back().empty()) : true)
)
||
(c != delimiter && parts.empty())
) {
parts.push_back({});
}
if (c != delimiter) {
parts.back().push_back(c);
}
}
if (skipEmpty && !parts.empty() && parts.back().empty()) {
parts.pop_back();
}
return parts;
}
void test(const std::string& line) {
std::cout << line << std::endl;
std::cout << "skipEmpty=0 |";
for (const std::string& part : split(line, ':')) {
std::cout << part << '|';
}
std::cout << std::endl;
std::cout << "skipEmpty=1 |";
for (const std::string& part : split(line, ':', true)) {
std::cout << part << '|';
}
std::cout << std::endl;
std::cout << std::endl;
}
int main() {
test("foo:bar:::baz");
test("");
test("foo");
test(":");
test("::");
test(":foo");
test("::foo");
test(":foo:");
test(":foo::");
return 0;
}
输出：
foo:bar:::baz
skipEmpty=0 |foo|bar|||baz|
skipEmpty=1 |foo|bar|baz|
skipEmpty=0 |
skipEmpty=1 |
foo
skipEmpty=0 |foo|
skipEmpty=1 |foo|
:
skipEmpty=0 |||
skipEmpty=1 |
::
skipEmpty=0 ||||
skipEmpty=1 |
:foo
skipEmpty=0 ||foo|
skipEmpty=1 |foo|
::foo
skipEmpty=0 |||foo|
skipEmpty=1 |foo|
:foo:
skipEmpty=0 ||foo||
skipEmpty=1 |foo|
:foo::
skipEmpty=0 ||foo|||
skipEmpty=1 |foo|
// adapted from a "regular" csv parse
std::string stringIn = "my csv is 10233478 NOTseparated by commas";
std::vector<std::string> commaSeparated(1);
int commaCounter = 0;
for (int i=0; i<stringIn.size(); i++) {
if (stringIn[i] == " ") {
commaSeparated.push_back("");
commaCounter++;
} else {
commaSeparated.at(commaCounter) += stringIn[i];
}
}
最后你会得到一个字符串向量，句子中的每个元素都用空格隔开。唯一的非标准资源是std:：vector（但由于涉及std:：string，我认为它是可以接受的）。
空字符串保存为单独的项。
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;
}
我编写了以下代码。您可以指定分隔符，它可以是字符串。结果类似于Java的String.split，结果中包含空字符串。
例如，如果我们调用split（“ABCPICKABCANYABCTWO:ABC”，“ABC”），结果如下：
0 <len:0>
1 PICK <len:4>
2 ANY <len:3>
3 TWO: <len:4>
4 <len:0>
代码：
vector <string> split(const string& str, const string& delimiter = " ") {
vector <string> tokens;
string::size_type lastPos = 0;
string::size_type pos = str.find(delimiter, lastPos);
while (string::npos != pos) {
// Found a token, add it to the vector.
cout << str.substr(lastPos, pos - lastPos) << endl;
tokens.push_back(str.substr(lastPos, pos - lastPos));
lastPos = pos + delimiter.size();
pos = str.find(delimiter, lastPos);
}
tokens.push_back(str.substr(lastPos, str.size() - lastPos));
return tokens;
}
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;
}
