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
C++20终于为我们提供了一个分裂函数。或者更确切地说，是一个范围适配器。螺栓连杆。
#include <iostream>
#include <ranges>
#include <string_view>
namespace ranges = std::ranges;
namespace views = std::views;
using str = std::string_view;
constexpr auto view =
"Multiple words"
| views::split(' ')
| views::transform([](auto &&r) -> str {
return {
&*r.begin(),
static_cast<str::size_type>(ranges::distance(r))
};
});
auto main() -> int {
for (str &&sv : view) {
std::cout << sv << '\n';
}
}
其他回答
#include <vector>
#include <string>
#include <sstream>
int main()
{
std::string str("Split me by whitespaces");
std::string buf; // Have a buffer string
std::stringstream ss(str); // Insert the string into a stream
std::vector<std::string> tokens; // Create vector to hold our words
while (ss >> buf)
tokens.push_back(buf);
return 0;
}
这是另一种方法。。
void split_string(string text,vector<string>& words)
{
int i=0;
char ch;
string word;
while(ch=text[i++])
{
if (isspace(ch))
{
if (!word.empty())
{
words.push_back(word);
}
word = "";
}
else
{
word += ch;
}
}
if (!word.empty())
{
words.push_back(word);
}
}
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());
}
我有两条线来解决这个问题：
char sep = ' ';
std::string s="1 This is an example";
for(size_t p=0, q=0; p!=s.npos; p=q)
std::cout << s.substr(p+(p!=0), (q=s.find(sep, p+1))-p-(p!=0)) << std::endl;
然后你可以把它放到一个向量中，而不是打印。
#include<iostream>
#include<string>
#include<sstream>
#include<vector>
using namespace std;
vector<string> split(const string &s, char delim) {
vector<string> elems;
stringstream ss(s);
string item;
while (getline(ss, item, delim)) {
elems.push_back(item);
}
return elems;
}
int main() {
vector<string> x = split("thi is an sample test",' ');
unsigned int i;
for(i=0;i<x.size();i++)
cout<<i<<":"<<x[i]<<endl;
return 0;
}
