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
虽然有一些答案提供了C++20解决方案，但自从发布以来，已经做了一些更改，并将其作为缺陷报告应用于C++20。正因为如此，解决方案变得更短、更好：
#include <iostream>
#include <ranges>
#include <string_view>
namespace views = std::views;
using str = std::string_view;
constexpr str text = "Lorem ipsum dolor sit amet, consectetur adipiscing elit.";
auto splitByWords(str input) {
return input
| views::split(' ')
| views::transform([](auto &&r) -> str {
return {r.begin(), r.end()};
});
}
auto main() -> int {
for (str &&word : splitByWords(text)) {
std::cout << word << '\n';
}
}
到今天为止，它仍然只在GCC的主干分支（Godbolt链接）上可用。它基于两个更改：P1391迭代器构造函数用于std:：string_view和P2210 DR修复std:：views:：split以保留范围类型。
在C++23中，不需要任何转换样板，因为P1989向std:：string_view:添加了一个范围构造函数
#include <iostream>
#include <ranges>
#include <string_view>
namespace views = std::views;
constexpr std::string_view text = "Lorem ipsum dolor sit amet, consectetur adipiscing elit.";
auto main() -> int {
for (std::string_view&& word : text | views::split(' ')) {
std::cout << word << '\n';
}
}
（螺栓连杆）
其他回答
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());
对于一个大得离谱而且可能是冗余的版本，可以尝试很多For循环。
string stringlist[10];
int count = 0;
for (int i = 0; i < sequence.length(); i++)
{
if (sequence[i] == ' ')
{
stringlist[count] = sequence.substr(0, i);
sequence.erase(0, i+1);
i = 0;
count++;
}
else if (i == sequence.length()-1) // Last word
{
stringlist[count] = sequence.substr(0, i+1);
}
}
它并不漂亮，但总的来说（除了标点符号和一系列其他错误）它是有效的！
有一个名为strtok的函数。
#include<string>
using namespace std;
vector<string> split(char* str,const char* delim)
{
char* saveptr;
char* token = strtok_r(str,delim,&saveptr);
vector<string> result;
while(token != NULL)
{
result.push_back(token);
token = strtok_r(NULL,delim,&saveptr);
}
return result;
}
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());
}
STL还没有这样的方法。
但是，您可以通过使用std:：string:：C_str（）成员来使用C的strtok（）函数，也可以编写自己的函数。下面是我在快速谷歌搜索（“STL字符串分割”）后找到的代码示例：
void Tokenize(const string& str,
vector<string>& tokens,
const string& delimiters = " ")
{
// Skip delimiters at beginning.
string::size_type lastPos = str.find_first_not_of(delimiters, 0);
// Find first "non-delimiter".
string::size_type pos = str.find_first_of(delimiters, lastPos);
while (string::npos != pos || string::npos != lastPos)
{
// Found a token, add it to the vector.
tokens.push_back(str.substr(lastPos, pos - lastPos));
// Skip delimiters. Note the "not_of"
lastPos = str.find_first_not_of(delimiters, pos);
// Find next "non-delimiter"
pos = str.find_first_of(delimiters, lastPos);
}
}
摘自：http://oopweb.com/CPP/Documents/CPPHOWTO/Volume/C++编程-HOWTO-7.html
如果您对代码示例有疑问，请留下评论，我会解释。
仅仅因为它没有实现称为迭代器的typedef或重载<<运算符，并不意味着它是错误的代码。我经常使用C函数。例如，printf和scanf都比std:：cin和std:：cout快（很明显），fopen语法对二进制类型更友好，它们也倾向于生成更小的EXE。
不要被这种“优雅胜过性能”的交易所吸引。
