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
这是我使用C++11和STL的解决方案。它应该是合理有效的：
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
std::vector<std::string> split(const std::string& s)
{
std::vector<std::string> v;
const auto end = s.end();
auto to = s.begin();
decltype(to) from;
while((from = std::find_if(to, end,
[](char c){ return !std::isspace(c); })) != end)
{
to = std::find_if(from, end, [](char c){ return std::isspace(c); });
v.emplace_back(from, to);
}
return v;
}
int main()
{
std::string s = "this is the string to split";
auto v = split(s);
for(auto&& s: v)
std::cout << s << '\n';
}
输出：
this
is
the
string
to
split
其他回答
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());
}
一些C++20编译器和大多数C++23编译器（range和string_view）
for (auto word : std::views::split("Somewhere down the road", ' '))
std::cout << std::string_view{ word.begin(), word.end() } << std::endl;
LazyString拆分器：
#include <string>
#include <algorithm>
#include <unordered_set>
using namespace std;
class LazyStringSplitter
{
string::const_iterator start, finish;
unordered_set<char> chop;
public:
// Empty Constructor
explicit LazyStringSplitter()
{}
explicit LazyStringSplitter (const string cstr, const string delims)
: start(cstr.begin())
, finish(cstr.end())
, chop(delims.begin(), delims.end())
{}
void operator () (const string cstr, const string delims)
{
chop.insert(delims.begin(), delims.end());
start = cstr.begin();
finish = cstr.end();
}
bool empty() const { return (start >= finish); }
string next()
{
// return empty string
// if ran out of characters
if (empty())
return string("");
auto runner = find_if(start, finish, [&](char c) {
return chop.count(c) == 1;
});
// construct next string
string ret(start, runner);
start = runner + 1;
// Never return empty string
// + tail recursion makes this method efficient
return !ret.empty() ? ret : next();
}
};
我将此方法称为LazyStringSplitter是因为一个原因——它不会一次性拆分字符串。本质上，它的行为类似于python生成器它公开了一个名为next的方法，该方法返回从原始字符串拆分的下一个字符串我使用了c++11STL中的无序集，因此查找分隔符的速度要快得多下面是它的工作原理
测试程序
#include <iostream>
using namespace std;
int main()
{
LazyStringSplitter splitter;
// split at the characters ' ', '!', '.', ','
splitter("This, is a string. And here is another string! Let's test and see how well this does.", " !.,");
while (!splitter.empty())
cout << splitter.next() << endl;
return 0;
}
输出，输出
This
is
a
string
And
here
is
another
string
Let's
test
and
see
how
well
this
does
改进这一点的下一个计划是实施开始和结束方法，以便可以执行以下操作：
vector<string> split_string(splitter.begin(), splitter.end());
#include <iostream>
#include <vector>
using namespace std;
int main() {
string str = "ABC AABCD CDDD RABC GHTTYU FR";
str += " "; //dirty hack: adding extra space to the end
vector<string> v;
for (int i=0; i<(int)str.size(); i++) {
int a, b;
a = i;
for (int j=i; j<(int)str.size(); j++) {
if (str[j] == ' ') {
b = j;
i = j;
break;
}
}
v.push_back(str.substr(a, b-a));
}
for (int i=0; i<v.size(); i++) {
cout<<v[i].size()<<" "<<v[i]<<endl;
}
return 0;
}
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;
}
};
