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
我的实施可以是另一种解决方案：
std::vector<std::wstring> SplitString(const std::wstring & String, const std::wstring & Seperator)
{
std::vector<std::wstring> Lines;
size_t stSearchPos = 0;
size_t stFoundPos;
while (stSearchPos < String.size() - 1)
{
stFoundPos = String.find(Seperator, stSearchPos);
stFoundPos = (stFoundPos == std::string::npos) ? String.size() : stFoundPos;
Lines.push_back(String.substr(stSearchPos, stFoundPos - stSearchPos));
stSearchPos = stFoundPos + Seperator.size();
}
return Lines;
}
测试代码：
std::wstring MyString(L"Part 1SEPsecond partSEPlast partSEPend");
std::vector<std::wstring> Parts = IniFile::SplitString(MyString, L"SEP");
std::wcout << L"The string: " << MyString << std::endl;
for (std::vector<std::wstring>::const_iterator it=Parts.begin(); it<Parts.end(); ++it)
{
std::wcout << *it << L"<---" << std::endl;
}
std::wcout << std::endl;
MyString = L"this,time,a,comma separated,string";
std::wcout << L"The string: " << MyString << std::endl;
Parts = IniFile::SplitString(MyString, L",");
for (std::vector<std::wstring>::const_iterator it=Parts.begin(); it<Parts.end(); ++it)
{
std::wcout << *it << L"<---" << std::endl;
}
测试代码的输出：
The string: Part 1SEPsecond partSEPlast partSEPend
Part 1<---
second part<---
last part<---
end<---
The string: this,time,a,comma separated,string
this<---
time<---
a<---
comma separated<---
string<---
其他回答
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。
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！
下面的代码使用strtok（）将字符串拆分为标记，并将标记存储在向量中。
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;
char one_line_string[] = "hello hi how are you nice weather we are having ok then bye";
char seps[] = " ,\t\n";
char *token;
int main()
{
vector<string> vec_String_Lines;
token = strtok( one_line_string, seps );
cout << "Extracting and storing data in a vector..\n\n\n";
while( token != NULL )
{
vec_String_Lines.push_back(token);
token = strtok( NULL, seps );
}
cout << "Displaying end result in vector line storage..\n\n";
for ( int i = 0; i < vec_String_Lines.size(); ++i)
cout << vec_String_Lines[i] << "\n";
cout << "\n\n\n";
return 0;
}
在getline上以“”作为标记进行循环。
这是我最喜欢的遍历字符串的方法。每个词你都可以做你想做的事。
string line = "a line of text to iterate through";
string word;
istringstream iss(line, istringstream::in);
while( iss >> word )
{
// Do something on `word` here...
}
