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
我刚刚写了一个很好的例子，说明如何按符号拆分一个字符，然后将每个字符数组（由符号分隔的单词）放入一个向量中。为了简单起见，我创建了std字符串的向量类型。
我希望这对你有帮助，并且对你可读。
#include <vector>
#include <string>
#include <iostream>
void push(std::vector<std::string> &WORDS, std::string &TMP){
WORDS.push_back(TMP);
TMP = "";
}
std::vector<std::string> mySplit(char STRING[]){
std::vector<std::string> words;
std::string s;
for(unsigned short i = 0; i < strlen(STRING); i++){
if(STRING[i] != ' '){
s += STRING[i];
}else{
push(words, s);
}
}
push(words, s);//Used to get last split
return words;
}
int main(){
char string[] = "My awesome string.";
std::cout << mySplit(string)[2];
std::cin.get();
return 0;
}
其他回答
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;
}
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<---
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！
另一种灵活快速的方式
template<typename Operator>
void tokenize(Operator& op, const char* input, const char* delimiters) {
const char* s = input;
const char* e = s;
while (*e != 0) {
e = s;
while (*e != 0 && strchr(delimiters, *e) == 0) ++e;
if (e - s > 0) {
op(s, e - s);
}
s = e + 1;
}
}
要将其与字符串向量一起使用（编辑：由于有人指出不继承STL类…hrmf；）：
template<class ContainerType>
class Appender {
public:
Appender(ContainerType& container) : container_(container) {;}
void operator() (const char* s, unsigned length) {
container_.push_back(std::string(s,length));
}
private:
ContainerType& container_;
};
std::vector<std::string> strVector;
Appender v(strVector);
tokenize(v, "A number of words to be tokenized", " \t");
就是这样！这只是使用tokenizer的一种方式，比如如何计数单词：
class WordCounter {
public:
WordCounter() : noOfWords(0) {}
void operator() (const char*, unsigned) {
++noOfWords;
}
unsigned noOfWords;
};
WordCounter wc;
tokenize(wc, "A number of words to be counted", " \t");
ASSERT( wc.noOfWords == 7 );
受限于想象力；）
