将字符串解析为int

Question

我对编码很陌生，我希望有人能帮助我?我试图在一行空格分隔的整数中读取，并将它们解析为(最终进入链表)向量。

所以一旦我有一个int型向量，有STL向量的迭代器，但我怎么能迭代通过链表中的节点不是STL?

#include <cstdlib>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
using namespace std;
int main(int argc, char** argv) {    
    cout << "Enter some integers, space delimited:n";
    string someString;
    getline(cin, someString);
    istringstream stringStream( someString );
    vector<string> parsedString;
    char splitToken = ' ';
    //read throguh the stream
    while(!stringstream.eof()){
        string subString;
        getline( stringStream, subString, splitToken);
        if(subString != ""){
        parsedString.push_back(subString);
        }
   }
    return EXIT_SUCCESS;
}

Answer 1

stringstream可以自动处理如下分隔符:

cout << "Enter some integers, space delimited:n";
string someString;
getline(cin, someString);
istringstream stringStream( someString );
vector<int> integers;
int n;
while (stringStream >> n)
    integers.push_back(n);

Answer 2

既然是空格分隔符，为什么不直接:

#include <iostream>
using namespace std;
int main() {
    int a;
    vector<int> v;
    while (cin >> a) {
        v.push_back(a);
    }
    for (int i = 0; i < v.size(); ++i) {
        int b = v[i];
    }
    return 0;
}

和BTW, ctrl-D或非整数输入，如char，将终止此while。