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);
}
最近我不得不将一个骆驼大小写的单词拆分成子单词。没有分隔符,只有大写字符。
#include <string>
#include <list>
#include <locale> // std::isupper
template<class String>
const std::list<String> split_camel_case_string(const String &s)
{
std::list<String> R;
String w;
for (String::const_iterator i = s.begin(); i < s.end(); ++i) { {
if (std::isupper(*i)) {
if (w.length()) {
R.push_back(w);
w.clear();
}
}
w += *i;
}
if (w.length())
R.push_back(w);
return R;
}
例如,这将“AQueryTrades”拆分为“A”、“Query”和“Trades”。该函数适用于窄字符串和宽字符串。因为它尊重当前的语言环境,所以将“RaumfahrtÜberwachungsVerordnung”分为“Raumfahrt”、“Überwachungs”和“Verordnug”。
注意std::upper应该真正作为函数模板参数传递。然后,此函数的更广义的from也可以在分隔符(如“、”、“;”或“”)处拆分。
#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|
我对string和u32string~的一般实现,使用boost::algorithm::split签名。
template<typename CharT, typename UnaryPredicate>
void split(std::vector<std::basic_string<CharT>>& split_result,
const std::basic_string<CharT>& s,
UnaryPredicate predicate)
{
using ST = std::basic_string<CharT>;
using std::swap;
std::vector<ST> tmp_result;
auto iter = s.cbegin(),
end_iter = s.cend();
while (true)
{
/**
* edge case: empty str -> push an empty str and exit.
*/
auto find_iter = find_if(iter, end_iter, predicate);
tmp_result.emplace_back(iter, find_iter);
if (find_iter == end_iter) { break; }
iter = ++find_iter;
}
swap(tmp_result, split_result);
}
template<typename CharT>
void split(std::vector<std::basic_string<CharT>>& split_result,
const std::basic_string<CharT>& s,
const std::basic_string<CharT>& char_candidate)
{
std::unordered_set<CharT> candidate_set(char_candidate.cbegin(),
char_candidate.cend());
auto predicate = [&candidate_set](const CharT& c) {
return candidate_set.count(c) > 0U;
};
return split(split_result, s, predicate);
}
template<typename CharT>
void split(std::vector<std::basic_string<CharT>>& split_result,
const std::basic_string<CharT>& s,
const CharT* literals)
{
return split(split_result, s, std::basic_string<CharT>(literals));
}