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
我们可以在c++中使用strtok，
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
char str[]="Mickey M;12034;911416313;M;01a;9001;NULL;0;13;12;0;CPP,C;MSC,3D;FEND,BEND,SEC;";
char *pch = strtok (str,";,");
while (pch != NULL)
{
cout<<pch<<"\n";
pch = strtok (NULL, ";,");
}
return 0;
}
其他回答
// 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。
#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|
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。
有一种更简单的方法可以做到这一点！！
#include <vector>
#include <string>
std::vector<std::string> splitby(std::string string, char splitter) {
int splits = 0;
std::vector<std::string> result = {};
std::string locresult = "";
for (unsigned int i = 0; i < string.size(); i++) {
if ((char)string.at(i) != splitter) {
locresult += string.at(i);
}
else {
result.push_back(locresult);
locresult = "";
}
}
if (splits == 0) {
result.push_back(locresult);
}
return result;
}
void printvector(std::vector<std::string> v) {
std::cout << '{';
for (unsigned int i = 0; i < v.size(); i++) {
if (i < v.size() - 1) {
std::cout << '"' << v.at(i) << "\",";
}
else {
std::cout << '"' << v.at(i) << "\"";
}
}
std::cout << "}\n";
}
#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <iterator>
#include <vector>
int main() {
using namespace std;
int n=8;
string sentence = "10 20 30 40 5 6 7 8";
istringstream iss(sentence);
vector<string> tokens;
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
back_inserter(tokens));
for(int i=0;i<n;i++){
cout<<tokens.at(i);
}
}
