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
值得一提的是，这里有另一种从输入字符串中提取令牌的方法，仅依赖于标准库设施。这是STL设计背后力量和优雅的一个例子。
#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <iterator>
int main() {
using namespace std;
string sentence = "And I feel fine...";
istringstream iss(sentence);
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
ostream_iterator<string>(cout, "\n"));
}
可以使用相同的通用复制算法将提取的令牌插入到容器中，而不是将其复制到输出流中。
vector<string> tokens;
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
back_inserter(tokens));
…或直接创建矢量：
vector<string> tokens{istream_iterator<string>{iss},
istream_iterator<string>{}};
其他回答
作为一个业余爱好者，这是我想到的第一个解决方案。我有点好奇，为什么我还没有在这里看到类似的解决方案，是不是我的做法有根本问题？
#include <iostream>
#include <string>
#include <vector>
std::vector<std::string> split(const std::string &s, const std::string &delims)
{
std::vector<std::string> result;
std::string::size_type pos = 0;
while (std::string::npos != (pos = s.find_first_not_of(delims, pos))) {
auto pos2 = s.find_first_of(delims, pos);
result.emplace_back(s.substr(pos, std::string::npos == pos2 ? pos2 : pos2 - pos));
pos = pos2;
}
return result;
}
int main()
{
std::string text{"And then I said: \"I don't get it, why would you even do that!?\""};
std::string delims{" :;\".,?!"};
auto words = split(text, delims);
std::cout << "\nSentence:\n " << text << "\n\nWords:";
for (const auto &w : words) {
std::cout << "\n " << w;
}
return 0;
}
http://cpp.sh/7wmzy
我使用以下代码：
namespace Core
{
typedef std::wstring String;
void SplitString(const Core::String& input, const Core::String& splitter, std::list<Core::String>& output)
{
if (splitter.empty())
{
throw std::invalid_argument(); // for example
}
std::list<Core::String> lines;
Core::String::size_type offset = 0;
for (;;)
{
Core::String::size_type splitterPos = input.find(splitter, offset);
if (splitterPos != Core::String::npos)
{
lines.push_back(input.substr(offset, splitterPos - offset));
offset = splitterPos + splitter.size();
}
else
{
lines.push_back(input.substr(offset));
break;
}
}
lines.swap(output);
}
}
// gtest:
class SplitStringTest: public testing::Test
{
};
TEST_F(SplitStringTest, EmptyStringAndSplitter)
{
std::list<Core::String> result;
ASSERT_ANY_THROW(Core::SplitString(Core::String(), Core::String(), result));
}
TEST_F(SplitStringTest, NonEmptyStringAndEmptySplitter)
{
std::list<Core::String> result;
ASSERT_ANY_THROW(Core::SplitString(L"xy", Core::String(), result));
}
TEST_F(SplitStringTest, EmptyStringAndNonEmptySplitter)
{
std::list<Core::String> result;
Core::SplitString(Core::String(), Core::String(L","), result);
ASSERT_EQ(1, result.size());
ASSERT_EQ(Core::String(), *result.begin());
}
TEST_F(SplitStringTest, OneCharSplitter)
{
std::list<Core::String> result;
Core::SplitString(L"x,y", L",", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(L"x", *result.begin());
ASSERT_EQ(L"y", *result.rbegin());
Core::SplitString(L",xy", L",", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(Core::String(), *result.begin());
ASSERT_EQ(L"xy", *result.rbegin());
Core::SplitString(L"xy,", L",", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(L"xy", *result.begin());
ASSERT_EQ(Core::String(), *result.rbegin());
}
TEST_F(SplitStringTest, TwoCharsSplitter)
{
std::list<Core::String> result;
Core::SplitString(L"x,.y,z", L",.", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(L"x", *result.begin());
ASSERT_EQ(L"y,z", *result.rbegin());
Core::SplitString(L"x,,y,z", L",,", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(L"x", *result.begin());
ASSERT_EQ(L"y,z", *result.rbegin());
}
TEST_F(SplitStringTest, RecursiveSplitter)
{
std::list<Core::String> result;
Core::SplitString(L",,,", L",,", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(Core::String(), *result.begin());
ASSERT_EQ(L",", *result.rbegin());
Core::SplitString(L",.,.,", L",.,", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(Core::String(), *result.begin());
ASSERT_EQ(L".,", *result.rbegin());
Core::SplitString(L"x,.,.,y", L",.,", result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(L"x", *result.begin());
ASSERT_EQ(L".,y", *result.rbegin());
Core::SplitString(L",.,,.,", L",.,", result);
ASSERT_EQ(3, result.size());
ASSERT_EQ(Core::String(), *result.begin());
ASSERT_EQ(Core::String(), *(++result.begin()));
ASSERT_EQ(Core::String(), *result.rbegin());
}
TEST_F(SplitStringTest, NullTerminators)
{
std::list<Core::String> result;
Core::SplitString(L"xy", Core::String(L"\0", 1), result);
ASSERT_EQ(1, result.size());
ASSERT_EQ(L"xy", *result.begin());
Core::SplitString(Core::String(L"x\0y", 3), Core::String(L"\0", 1), result);
ASSERT_EQ(2, result.size());
ASSERT_EQ(L"x", *result.begin());
ASSERT_EQ(L"y", *result.rbegin());
}
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;
}
};
使用std：：stringstream非常好，并且完全符合您的要求。如果您只是在寻找不同的方法，那么可以使用std:：find（）/std:：find_first_of（）和std:：string：：substr（）。
下面是一个示例：
#include <iostream>
#include <string>
int main()
{
std::string s("Somewhere down the road");
std::string::size_type prev_pos = 0, pos = 0;
while( (pos = s.find(' ', pos)) != std::string::npos )
{
std::string substring( s.substr(prev_pos, pos-prev_pos) );
std::cout << substring << '\n';
prev_pos = ++pos;
}
std::string substring( s.substr(prev_pos, pos-prev_pos) ); // Last word
std::cout << substring << '\n';
return 0;
}
值得一提的是，这里有另一种从输入字符串中提取令牌的方法，仅依赖于标准库设施。这是STL设计背后力量和优雅的一个例子。
#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <iterator>
int main() {
using namespace std;
string sentence = "And I feel fine...";
istringstream iss(sentence);
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
ostream_iterator<string>(cout, "\n"));
}
可以使用相同的通用复制算法将提取的令牌插入到容器中，而不是将其复制到输出流中。
vector<string> tokens;
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
back_inserter(tokens));
…或直接创建矢量：
vector<string> tokens{istream_iterator<string>{iss},
istream_iterator<string>{}};
