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);
}
这是我的版本获取了Kev的来源:
#include <string>
#include <vector>
void split(vector<string> &result, string str, char delim ) {
string tmp;
string::iterator i;
result.clear();
for(i = str.begin(); i <= str.end(); ++i) {
if((const char)*i != delim && i != str.end()) {
tmp += *i;
} else {
result.push_back(tmp);
tmp = "";
}
}
}
之后,调用函数并执行以下操作:
vector<string> hosts;
split(hosts, "192.168.1.2,192.168.1.3", ',');
for( size_t i = 0; i < hosts.size(); i++){
cout << "Connecting host : " << hosts.at(i) << "..." << endl;
}
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!