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
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！
其他回答
谢谢@Jairo Abdiel Toribio Cisneros。它对我有效，但您的函数返回一些空元素。因此，对于没有空的返回，我编辑了以下内容：
std::vector<std::string> split(std::string str, const char* delim) {
std::vector<std::string> v;
std::string tmp;
for(std::string::const_iterator i = str.begin(); i <= str.end(); ++i) {
if(*i != *delim && i != str.end()) {
tmp += *i;
} else {
if (tmp.length() > 0) {
v.push_back(tmp);
}
tmp = "";
}
}
return v;
}
使用：
std::string s = "one:two::three";
std::string delim = ":";
std::vector<std::string> vv = split(s, delim.c_str());
这是我写的一个函数，帮助我做了很多事情。它在为WebSocket做协议时帮助了我。
using namespace std;
#include <iostream>
#include <vector>
#include <sstream>
#include <string>
vector<string> split ( string input , string split_id ) {
vector<string> result;
int i = 0;
bool add;
string temp;
stringstream ss;
size_t found;
string real;
int r = 0;
while ( i != input.length() ) {
add = false;
ss << input.at(i);
temp = ss.str();
found = temp.find(split_id);
if ( found != string::npos ) {
add = true;
real.append ( temp , 0 , found );
} else if ( r > 0 && ( i+1 ) == input.length() ) {
add = true;
real.append ( temp , 0 , found );
}
if ( add ) {
result.push_back(real);
ss.str(string());
ss.clear();
temp.clear();
real.clear();
r = 0;
}
i++;
r++;
}
return result;
}
int main() {
string s = "S,o,m,e,w,h,e,r,e, down the road \n In a really big C++ house. \n Lives a little old lady. \n That no one ever knew. \n She comes outside. \n In the very hot sun. \n\n\n\n\n\n\n\n And throws C++ at us. \n The End. FIN.";
vector < string > Token;
Token = split ( s , "," );
for ( int i = 0 ; i < Token.size(); i++) cout << Token.at(i) << endl;
cout << endl << Token.size();
int a;
cin >> a;
return a;
}
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！
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<---
void splitString(string str, char delim, string array[], const int arraySize)
{
int delimPosition, subStrSize, subStrStart = 0;
for (int index = 0; delimPosition != -1; index++)
{
delimPosition = str.find(delim, subStrStart);
subStrSize = delimPosition - subStrStart;
array[index] = str.substr(subStrStart, subStrSize);
subStrStart =+ (delimPosition + 1);
}
}
